Week 3 Notes

Mathematics for
Data Science 2

Vector spaces and what makes them tick: the nine axioms, their consequences, subspaces, linear combinations, linear dependence and independence, and spanning sets.

10 chapters78 worked examples~73 min read

AxiomsSubspacesIndependenceSpan

Chapters

01Why Vector Spaces? A Motivating SystemPart 1
02The Vector Space AxiomsPart 2
03Is It a Vector Space? Worked ExamplesPart 3
04Properties of Vector SpacesPart 4
05SubspacesPart 5
06Linear CombinationsPart 6
07Linear Dependence and IndependencePart 7
08Checking Independence With MatricesPart 8
09Span and Spanning SetsPart 9
10Week 3 Summary SheetAll parts

Foundations

Why Vector Spaces? A Motivating System

Solution sets of homogeneous systems are closed under addition and scaling. That one property is what a vector space abstracts.

1.1A motivating system

The last two weeks were about solving systems of linear equations. Vector spaces give a bird's-eye view of what the solutions look like as a whole. (The idea was formalised by Giuseppe Peano in 1888.) Start with one homogeneous system:

x + y - 2z = 0,\qquad x - 2y + z = 0,\qquad -2x + y + z = 0

In matrix form, with $A = \begin{bmatrix}1&1&-2\\1&-2&1\\-2&1&1\end{bmatrix}$ , the system is $A\mathbf{x} = \mathbf{0}$ . Solving it means finding $x, y, z$ so that a combination of the columns of $A$ gives zero. Collect every solution into one set:

S := \left\{ \begin{bmatrix}a\\b\\c\end{bmatrix} \in \mathbb{R}^{3\times 1} \;\middle|\; A\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} \right\}

1.2The key observation: closure

S is closed under both operations

If $v_1, v_2 \in S$ then $v_1 + v_2 \in S$ , because $A(v_1 + v_2) = Av_1 + Av_2 = 0 + 0 = 0$ .
If $v_1 \in S$ then $\alpha v_1 \in S$ for all real $\alpha$ , because $A(\alpha v_1) = \alpha(Av_1) = \alpha \cdot 0 = 0$ .

Together: if $v_1, v_2 \in S$ , then $\alpha_1 v_1 + \alpha_2 v_2 \in S$ for all real $\alpha_1, \alpha_2$ . The expression $\alpha_1 v_1 + \alpha_2 v_2$ is called a linear combination of $v_1$ and $v_2$ . More generally, a combination of any $v_1, \dots, v_n \in S$ stays in $S$ .

1.3Seeing S concretely

Example 1Solve the motivating system

Find the set $S$ of all solutions of $x + y - 2z = 0,\ x - 2y + z = 0,\ -2x + y + z = 0$ .

Solution

Row-reduce $A$ : $R_2 - R_1$ and $R_3 + 2R_1$ give $\begin{bmatrix}1&1&-2\\0&-3&3\\0&3&-3\end{bmatrix}$ ; then $R_3 + R_2$ gives $\begin{bmatrix}1&1&-2\\0&-3&3\\0&0&0\end{bmatrix}$ .
Row 2: $-3y + 3z = 0$ , so $y = z$ . Row 1: $x = -y + 2z = z$ .
$z$ is free; put $z = t$ . Then $(x, y, z) = (t, t, t)$ .
$S = \{\, t\,(1, 1, 1)^T : t \in \mathbb{R} \,\}$ , a line through the origin.
Check: $1 + 1 - 2 = 0$ , $1 - 2 + 1 = 0$ , $-2 + 1 + 1 = 0$ . Adding two points $t_1(1,1,1) + t_2(1,1,1) = (t_1 + t_2)(1,1,1)$ stays on the line, which is exactly the closure above.

Figure 1. Points of the solution line S stay on S when added or scaled

1.4What is a vector space, roughly?

Roughly, a vector space $V$ is any collection of objects that behaves like $\mathbb{R}^n$ : you can add elements of $V$ , and multiply elements of $V$ by real scalars, and these two operations obey a short list of rules called the axioms (Part 2).

1.5A contrast: the non-homogeneous case

Example 2When the right-hand side is not zero

Let $T$ be the solution set of $x + 2y = 4$ in $\mathbb{R}^2$ , and take $u = (4, 0)$ , $v = (0, 2)$ . Are $u, v \in T$ ? Is $u + v \in T$ ? Is $2u \in T$ ? What equation does $u - v$ satisfy?

Solution

$4 + 2(0) = 4$ and $0 + 2(2) = 4$ , so $u, v \in T$ .
$u + v = (4, 2)$ and $4 + 2(2) = 8 \neq 4$ , so $u + v \notin T$ .
$2u = (8, 0)$ and $8 + 0 = 8 \neq 4$ , so $2u \notin T$ .
$u - v = (4, -2)$ and $4 + 2(-2) = 0$ : it solves the homogeneous equation $x + 2y = 0$ , not the original.
In general, if $Au = b$ and $Av = b$ , then $A(u + v) = 2b$ , $A(cu) = cb$ and $A(u - v) = 0$ . With $b \neq 0$ : $2b \neq b$ and $cb \neq b$ (for $c \neq 1$ ), so $T$ is not closed. The right-hand side being zero is what makes $S$ behave like a vector space.

Example 3A plane of solutions

Consider $x + 2y - z = 0,\ 2x + 4y - 2z = 0$ . Find the solution set $S$ and show that $3(1, 0, 1) - 2(-2, 1, 0) \in S$ .

Solution

$R_2 - 2R_1$ gives $\begin{bmatrix}1&2&-1\\0&0&0\end{bmatrix}$ : the second equation is just twice the first, so one equation survives, $x + 2y - z = 0$ .
The only pivot is in the $x$ -column, so $y$ and $z$ are free: $y = s$ , $z = t$ , $x = -2s + t$ .
$S = \{ s(-2, 1, 0) + t(1, 0, 1) : s, t \in \mathbb{R} \}$ , a plane through the origin in $\mathbb{R}^3$ .
$(1, 0, 1)$ : $1 + 0 - 1 = 0$ $\checkmark$ . $(-2, 1, 0)$ : $-2 + 2 - 0 = 0$ $\checkmark$ .
$3(1, 0, 1) - 2(-2, 1, 0) = (3, 0, 3) + (4, -2, 0) = (7, -2, 3)$ , and $7 + 2(-2) - 3 = 0$ $\checkmark$ , so it is in $S$ .

1.6Going deeper

Why this has to work. For any $v_1, v_2 \in S$ and scalars $\alpha, \beta$ : $A(\alpha v_1 + \beta v_2) = \alpha Av_1 + \beta Av_2 = 0$ . Every combination of solutions is again a solution.
Lines and planes through the origin. Solution sets of homogeneous systems are lines, planes (or all of $\mathbb{R}^n$ , or just $\{0\}$ ) through the origin. A set that misses the origin can never behave like this (extra).
Always a solution. $\mathbf{x} = \mathbf{0}$ always solves $A\mathbf{x} = \mathbf{0}$ , so $S$ is never empty.

1.7Exam tips and common mistakes

Exam tips

To test whether a solution set is "nice", plug the sum and a scalar multiple back into the equation.
A homogeneous system always contains the zero vector; a non-homogeneous one never does.

Common mistakes

Assuming that every solution set is closed. Only the homogeneous case is.
Forgetting that $x = -2s + t$ depends on both free variables.

Practice questions (graded-assignment style)

Q1The solution set

Multiple Choice Question (MCQ): choose the one correct option.

The solution set of $x + y - 2z = 0,\ x - 2y + z = 0,\ -2x + y + z = 0$ is:

Only the zero vector

A line through the origin,

\{t(1, 1, 1)\}

A plane through the origin

Empty

Solution

Idea

Row-reduce and count the free variables.

$\operatorname{rank}A = 2$ with 3 unknowns, so there is exactly one free variable.
$y = z$ and $x = z$ , so every solution is $t(1, 1, 1)$ : a line through the origin.
Answer: option 2.

Q2Which vectors lie in S?

Multiple Select Question (MSQ): select all that apply.

For $x + 2y - z = 0,\ 2x + 4y - 2z = 0$ , choose the vectors that belong to the solution set.

(1, 0, 1)

(-2, 1, 0)

(7, -2, 3)

(1, 1, 1)

Solution

Idea

Only the first equation matters; the second is twice it.

$(1, 0, 1)$ : $1 + 0 - 1 = 0$ . $(-2, 1, 0)$ : $-2 + 2 - 0 = 0$ . $(7, -2, 3)$ : $7 - 4 - 3 = 0$ .
$(1, 1, 1)$ : $1 + 2 - 1 = 2 \neq 0$ , so it is not in $S$ .
Answer: options 1, 2 and 3.

Q3The non-homogeneous set T

Multiple Select Question (MSQ): select all that apply.

$T = \{(x, y) : x + 2y = 4\}$ , $u = (4, 0)$ , $v = (0, 2)$ . Choose the correct statements.

u, v \in T

u + v \in T

2u \notin T

u - v

satisfies

x + 2y = 0

Solution

Idea

Plug each combination into $x + 2y$ .

$u, v$ give 4 and 4, so both are in $T$ (option 1 true).
$u + v = (4, 2)$ gives 8, so it is not in $T$ (option 2 false).
$2u = (8, 0)$ gives 8, so $2u \notin T$ (option 3 true).
$u - v = (4, -2)$ gives 0 (option 4 true).
Answer: options 1, 3 and 4.

Q4A point of the line

Numerical Answer Type (NAT).

A point $(x, y, z)$ of the solution line $S = \{t(1, 1, 1)\}$ has $x = 5$ . Find $x + y + z$ .

Solution

Idea

Every point of S is (t, t, t).

$x = t = 5$ , so the point is $(5, 5, 5)$ .
$x + y + z = 15$ .
Answer: 15

Q5Closed under both operations?

Multiple Choice Question (MCQ): choose the one correct option.

Which of these subsets of $\mathbb{R}^2$ is closed under addition and under multiplication by every real number?

x + y = 1

x + y = 0

x^2 + y^2 = 1

y \ge 0

Solution

Idea

Test with a sum and with the scalar −1.

$x + y = 0$ : $(a, -a) + (b, -b) = (a + b, -(a + b))$ and $c(a, -a) = (ca, -ca)$ , both on the line.
$x + y = 1$ fails (sum gives 2), $x^2 + y^2 = 1$ fails (sum of $(1,0)$ and $(0,1)$ has length $\sqrt2$ ), $y \ge 0$ fails (times $-1$ ).
Answer: option 2.

Revision overview

Homogeneous systemAx = 0
x = 0 is always a solution

Solution set SClosed under + and ×
Line or plane through O

Linear combinationα₁v₁ + α₂v₂ + … + αₙvₙ
of solutions is a solution

Why vector spaces?Week 3 · Part 1

ContrastAx = b, b ≠ 0
is not closed

IdeaA vector space: add and
scale, obeying axioms

NextThe nine axioms

Revision: why solution sets of homogeneous systems motivate vector spaces

Next · FoundationsThe Vector Space Axioms

Foundations

The Vector Space Axioms

The nine rules that define a vector space, and how a single failed rule rules a set out.

2.1The definition

Definition: vector space

A vector space $(V, +, \cdot, \mathbb{R})$ is a non-empty set $V$ together with two operations

\begin{aligned}+ \ (\text{addition}) &:\ V \times V \to V\\ \cdot \ (\text{scalar multiplication}) &:\ \mathbb{R} \times V \to V\end{aligned}

which satisfy the nine conditions below for all $u, v, w \in V$ and all scalars $a, b \in \mathbb{R}$ .

#	Axiom	Statement
1	Additive closure	$u + v \in V$
2	Multiplicative closure	$a \cdot u \in V$
3	Additive commutativity	$u + v = v + u$
4	Additive associativity	$(u + v) + w = u + (v + w)$
5	Existence of a zero vector	there is $0_V \in V$ with $u + 0_V = u$
6	Additive inverse	for every $u$ there is $u\prime \in V$ with $u + u\prime = 0_V$
7	Unity	$1 \cdot u = u$
8	Associativity of multiplication	$(ab) \cdot u = a \cdot (b \cdot u)$
9	Distributivity	$(a + b) \cdot u = a \cdot u + b \cdot u$ and $a \cdot (u + v) = a \cdot u + a \cdot v$

Figure 2. The nine axioms in three groups

2.2Things to keep in mind

If even one axiom fails, $V$ is not a vector space. To prove that something is not a vector space, one counterexample to one axiom is enough.
The zero element $0_V$ of a vector space is always unique (proved in Part 4).
The real number $0$ and the zero vector $0_V$ are both usually written $0$ ; the context tells you which is meant.
We drop the dot and write $au$ instead of $a \cdot u$ .
A vector is simply an element of a vector space. It need not be an arrow or a tuple: matrices, and even pairs $(x, 1)$ , are vectors once they live in a vector space.
The zero vector is whatever satisfies axiom 5 for the given operations. It is not automatically $(0, 0, \dots, 0)$ . In Example 12 below the zero vector is $(-1, 0)$ .

2.3Exam tips and common mistakes

Exam tips

Read the operations carefully: they may not be the usual ones.
First look for a zero vector and then test unity and distributivity: they fail most often.
To show "not a vector space", write one concrete counterexample with numbers.

Common mistakes

Assuming the zero vector is the tuple of zeros.
Checking a few axioms and declaring "vector space". All nine must be verified.
Mixing up the two distributive laws. Both must hold.

Practice questions (graded-assignment style)

Q1Which axiom fails?

Multiple Choice Question (MCQ): choose the one correct option.

On $\mathbb{R}^2$ with the usual addition define $c \odot (x, y) = (cx,\ y)$ . Which axiom fails?

Axiom 7 (unity)

Axiom 8 (associativity of multiplication)

Axiom 9, first law

(a + b) \odot u = a \odot u + b \odot u

Axiom 2 (multiplicative closure)

Solution

Idea

Test each law on u = (0, 1) or (1, 1).

Unity: $1 \odot (x, y) = (x, y)$ holds. Axiom 8: $(ab)x = a(bx)$ holds. Closure holds.
First law: $(a + b) \odot (x, y) = ((a + b)x,\ y)$ , but $a \odot u + b \odot u = (ax + bx,\ 2y)$ . They differ whenever $y \neq 0$ .
Answer: option 3.

Q2Facts about the axioms

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements.

If one axiom fails, the set is not a vector space.

The zero vector of a vector space is always

(0, 0, \dots, 0)

A vector must be an arrow or a tuple.

The zero vector of a vector space is unique.

Solution

Idea

Recall "things to keep in mind".

Statement 1 and 4 are true.
Statement 2 is false: in Example 12 the zero vector is $(-1, 0)$ .
Statement 3 is false: matrices, pairs and numbers can all be vectors.
Answer: options 1 and 4.

Q3Addition-only axioms

Numerical Answer Type (NAT).

How many of the nine axioms involve only the addition operation (no scalar)?

Solution

Idea

List the axioms by number.

Axioms 1, 3, 4, 5 and 6 involve only addition.
Axioms 2, 7, 8, 9 involve scalar multiplication. So the answer is 5.
Answer: 5

Q4An unusual scalar action

Numerical Answer Type (NAT).

With $c \odot (x, y) = (c^2 x,\ c^2 y)$ , compute the first entry of $2 \odot (1, 0)$ .

Solution

Idea

Substitute c = 2.

$2 \odot (1, 0) = (2^2 \cdot 1,\ 2^2 \cdot 0) = (4, 0)$ , so the first entry is 4.
Compare $1 \odot (1,0) + 1 \odot (1,0) = (2, 0)$ : the first distributive law fails.
Answer: 4

Q5Axioms that hold

Multiple Select Question (MSQ): select all that apply.

For $c \odot (x, y) = (c^2 x,\ c^2 y)$ on $\mathbb{R}^2$ with the usual addition, which axioms hold?

Axiom 7 (unity)

Axiom 8 (associativity of multiplication)

Distributivity

a \odot (u + v) = a \odot u + a \odot v

Distributivity

(a + b) \odot u = a \odot u + b \odot u

Solution

Idea

Compute both sides of each law.

Unity: $1^2 = 1$ . Axiom 8: $(ab)^2 = a^2 b^2$ . Second law: $a^2(x_1 + x_2) = a^2 x_1 + a^2 x_2$ .
First law fails: $(a + b)^2 \neq a^2 + b^2$ in general.
Answer: options 1, 2 and 3.

Revision overview

Closure1: u + v ∈ V
2: a·u ∈ V

Addition rules3 commutative, 4 associative
5 zero vector, 6 inverse

Scalar rules7 unity, 8 (ab)u = a(bu)
9 two distributive laws

Vector space axiomsWeek 3 · Part 2

To disproveOne counterexample to one
axiom is enough

Zero vectorWhatever satisfies axiom 5
Not always (0, …, 0)

VectorAny element of a
vector space

Revision: what the nine axioms say and how to break them

Next · FoundationsIs It a Vector Space? Worked Examples

Foundations

Is It a Vector Space? Worked Examples

Standard spaces and strange ones: the same recipe decides every case.

The recipe is always the same: read the operations carefully (they may not be the usual ones!), then check the nine axioms one by one, or hunt for one that fails.

3.1The standard examples

Example 7 $\mathbb{R}^n$ with the usual operations

Show that $V = \mathbb{R}^n$ with $(x_1, \dots, x_n) + (y_1, \dots, y_n) = (x_1 + y_1, \dots, x_n + y_n)$ and $c\cdot(x_1, \dots, x_n) = (cx_1, \dots, cx_n)$ is a vector space.

Solution

Take $u = (x_i)$ , $v = (y_i)$ , $w = (z_i)$ . The operations inside the brackets are ordinary real operations.

1. Additive closure. $u + v = (x_i + y_i)$ is an $n$ -tuple of reals, so it is in $V$ .
2. Multiplicative closure. $c\cdot u = (cx_i) \in V$ for every $c$ .
3. Commutativity. $x_i + y_i = y_i + x_i$ for each $i$ because real addition commutes.
4. Associativity. $(x_i + y_i) + z_i = x_i + (y_i + z_i)$ for each $i$ .
5. Zero vector. $u + 0_V = u$ forces $x_i + m_i = x_i$ , so $m_i = 0$ and $0_V = (0, \dots, 0)$ .
6. Inverse. $x_i + a_i = 0$ gives $a_i = -x_i$ , so $u\prime = (-x_1, \dots, -x_n)$ .
7. Unity. $1\cdot u = (1x_i) = u$ .
8. $(ab)\cdot u = ((ab)x_i) = (a(bx_i)) = a\cdot(b\cdot u)$ .
9. Distributivity. $(a + b)x_i = ax_i + bx_i$ and $a(x_i + y_i) = ax_i + ay_i$ for each $i$ .

All nine hold, so $\mathbb{R}^n$ is a vector space.

Example 8The matrices $M_{2\times 3}(\mathbb{R})$

Show that the set of all $2 \times 3$ real matrices with entry-wise addition and scalar multiplication is a vector space.

Solution

Adding two $2\times3$ matrices entry by entry, or multiplying every entry by $c$ , gives another $2 \times 3$ real matrix: axioms 1 and 2 hold.
Zero vector: the zero matrix $\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}$ , since adding it changes no entry.
Inverse: for $u = (a_{ij})$ take $-u = (-a_{ij})$ ; then $u + (-u)$ is the zero matrix.
The remaining axioms work entry by entry exactly as in $\mathbb{R}^n$ : a $2 \times 3$ matrix is just 6 real numbers in a grid. So it is a vector space.

Example 9Solutions of $A\mathbf{x} = \mathbf{0}$

Let $A$ be a real $m \times n$ matrix and $V = \{x \in \mathbb{R}^n : Ax = 0\}$ with the usual column-vector operations. Show $V$ is a vector space.

Solution

Additive closure. If $Au = 0$ and $Av = 0$ , then $A(u + v) = Au + Av = 0 + 0 = 0$ , so $u + v \in V$ .
Multiplicative closure. $A(cu) = c(Au) = c \cdot 0 = 0$ , so $cu \in V$ .
Zero and inverses. The zero column satisfies $A0 = 0$ , so it is in $V$ ; and if $Au = 0$ then $A(-u) = -Au = 0$ , so $-u \in V$ .
The other axioms are inherited from $\mathbb{R}^n$ . Hence $V$ is a vector space: the general version of the set $S$ of Part 1.

3.2Strange operations

Example 10A strange addition (fails)

On $V = \mathbb{R}^2$ define $(x_1, x_2) + (y_1, y_2) = (x_1 + y_1,\ x_2 - y_2)$ and $c\cdot(x_1, x_2) = (cx_1, cx_2)$ . Is it a vector space?

Solution

Closure holds: $(x_1 + y_1, x_2 - y_2)$ and $(cx_1, cx_2)$ are both in $\mathbb{R}^2$ .
Commutativity fails: $(1, 1) + (2, 3) = (3, -2)$ , but $(2, 3) + (1, 1) = (3, 2)$ . In general the second components $x_2 - y_2$ and $y_2 - x_2$ differ whenever $x_2 \neq y_2$ . Another check: $(4, 1) + (1, 4) = (5, -3)$ .
One axiom fails, so it is not a vector space. (Several others fail too, associativity for example.)

Example 11The set $\{(x, 1)\}$

Let $V = \{(x, 1) : x \in \mathbb{R}\}$ with $(x, 1) + (y, 1) := (x + y,\ 1)$ and $c\cdot(x, 1) := (cx,\ 1)$ . Is it a vector space?

Solution

1, 2. $(x + y, 1)$ and $(cx, 1)$ both have second entry 1, so they are in $V$ .
3, 4. $(x + y, 1) = (y + x, 1)$ ; $((x + y) + z, 1) = (x + (y + z), 1)$ .
5. Zero vector. Want $0_V = (m, 1)$ with $(x, 1) + (m, 1) = (x, 1)$ , i.e. $(x + m, 1) = (x, 1)$ , so $m = 0$ and $0_V = (0, 1)$ .
6. Inverse. Want $(a, 1)$ with $(x + a, 1) = (0, 1)$ : $a = -x$ , so $u\prime = (-x, 1)$ .
7. $1\cdot(x, 1) = (x, 1)$ . 8. $(ab)\cdot u = ((ab)x, 1) = a\cdot(bx, 1) = a\cdot(b\cdot u)$ .
9. $(a + b)\cdot u = ((a + b)x, 1) = (ax, 1) + (bx, 1)$ and $a\cdot(u + v) = (ax + ay, 1) = a\cdot u + a\cdot v$ .
All nine hold, so it is a vector space, even though $(0, 0)$ is not in it. Its zero vector is $(0, 1)$ .

Example 12Shifted operations on the plane

On $V = \mathbb{R}^2$ define $(x_1, y_1) + (x_2, y_2) := (x_1 + x_2 + 1,\ y_1 + y_2)$ and $c\cdot(x, y) := (cx + c - 1,\ cy)$ . Show that $V$ is a vector space and find its zero vector.

Solution

Examples: $(1, 1) + (2, 1) = (4, 2)$ ; $(2, 1) + (-2, 3) = (1, 4)$ ; $4\cdot(1, 1) = (4 + 4 - 1,\ 4) = (7, 4)$ .
Zero vector. Want $0_V = (m, n)$ with $(x, y) + (m, n) = (x, y)$ : $(x + m + 1,\ y + n) = (x, y)$ gives $m = -1$ , $n = 0$ . So $0_V = (-1, 0)$ .
Inverse. Want $(a, b)$ with $(x + a + 1,\ y + b) = (-1, 0)$ : $a = -x - 2$ , $b = -y$ . The inverse of $(x, y)$ is $(-x - 2,\ -y)$ . Check: $(x - x - 2 + 1,\ 0) = (-1, 0) = 0_V$ . $\checkmark$
Unity. $1\cdot(x, y) = (x + 1 - 1,\ y) = (x, y)$ .
Axiom 8. $a\cdot(b\cdot u) = a\cdot(bx + b - 1,\ by) = (a(bx + b - 1) + a - 1,\ aby) = (abx + ab - 1,\ aby) = (ab)\cdot u$ .
Distributivity. $a\cdot u + b\cdot u = (ax_1 + a - 1,\ ay_1) + (bx_1 + b - 1,\ by_1) = ((a + b)x_1 + (a + b) - 1,\ (a + b)y_1) = (a + b)\cdot u$ . For $a\cdot(u + v)$ : both sides have first entry $ax_1 + ax_2 + 2a - 1$ .
Closure, commutativity and associativity of addition are immediate. Hence $V$ is a vector space.

Common slip. Solving $x + a + 1 = 0$ as if the zero vector were $(0, 0)$ gives $(-x - 1, -y)$ , but $(x, y) + (-x - 1, -y) = (0, 0)$ , which is not $0_V$ here. The inverse condition is always $u + u\prime = 0_V$ with the zero vector you actually found.

Why it works. Relabel every point by shifting one unit right: $\varphi(x, y) = (x + 1, y)$ . Then $\varphi(u + v) = \varphi(u) + \varphi(v)$ and $\varphi(c\cdot u) = c\,\varphi(u)$ with the usual operations. So this is ordinary $\mathbb{R}^2$ with its origin moved to $(-1, 0)$ , which is exactly where the zero vector landed.

Figure 3. The shifted plane: same points, but the zero vector is (−1, 0)

Example 13Matrices with fixed 1s on the diagonal

Let $V = \left\{\begin{bmatrix}1&a\\b&1\end{bmatrix} : a, b \in \mathbb{R}\right\}$ with $\begin{bmatrix}1&a\\b&1\end{bmatrix} + \begin{bmatrix}1&c\\d&1\end{bmatrix} = \begin{bmatrix}1&a + c\\b + d&1\end{bmatrix}$ and $p\cdot\begin{bmatrix}1&a\\b&1\end{bmatrix} = \begin{bmatrix}1&pa\\pb&1\end{bmatrix}$ . Is $V$ a vector space?

Solution

This "addition" is not ordinary matrix addition (that would put 2s on the diagonal). The operations act only on the off-diagonal pair $(a, b)$ ; the diagonal 1s stay fixed.
Closure, commutativity, associativity. Results have 1s on the diagonal and reals off it, so they are in $V$ ; off-diagonal entries add as real numbers, so axioms 3 and 4 hold.
Zero vector. Try $0_V = \begin{bmatrix}1&m\\n&1\end{bmatrix}$ : $a + m = a$ and $b + n = b$ give $m = n = 0$ , so $0_V = I_2$ , the identity matrix (the zero matrix is not even in $V$ ).
Inverse. Need $a + a\prime = 0$ and $b + b\prime = 0$ : $u\prime = \begin{bmatrix}1&-a\\-b&1\end{bmatrix}$ .
7, 8. $1\cdot u = u$ and $(pq)\cdot u = p\cdot(q\cdot u)$ since $pq\,a = p(qa)$ .
9. $(p + q)\cdot u = p\cdot u + q\cdot u$ and $p\cdot(u + v) = p\cdot u + p\cdot v$ , by distributing on the off-diagonal entries.
All nine hold: yes, $V$ is a vector space. It is a copy of $\mathbb{R}^2$ with $(a, b)$ stored off the diagonal.

Example 14Positive reals with multiplication as "addition"

Let $V = \{x \in \mathbb{R} : x > 0\}$ with $x \oplus y = xy$ and $c \odot x = x^c$ . Show that $V$ is a vector space. What is its zero vector, and what is the additive inverse of 5?

Solution

Closure. If $x, y > 0$ then $xy > 0$ ; if $x > 0$ then $x^c > 0$ for every real $c$ .
3, 4. $x \oplus y = xy = yx = y \oplus x$ and $(x \oplus y) \oplus z = (xy)z = x(yz) = x \oplus (y \oplus z)$ .
Zero vector. Need $m$ with $x \oplus m = x$ for all $x$ : $xm = x$ , so $m = 1$ . Thus $0_V = 1$ .
Inverse. Need $x\prime$ with $x \oplus x\prime = 1$ : $xx\prime = 1$ , so $x\prime = 1/x$ (positive, so in $V$ ). The inverse of 5 is $\tfrac15$ : $5 \oplus \tfrac15 = 5\cdot\tfrac15 = 1 = 0_V$ . $\checkmark$
7, 8. $1 \odot x = x^1 = x$ ; $(ab) \odot x = x^{ab} = (x^b)^a = a \odot (b \odot x)$ .
9. $(a + b) \odot x = x^{a+b} = x^a x^b = (a \odot x) \oplus (b \odot x)$ and $a \odot (x \oplus y) = (xy)^a = x^a y^a = (a \odot x) \oplus (a \odot y)$ .
All nine hold, so $V$ is a vector space. Sanity check: $0 \odot x = x^0 = 1 = 0_V$ and $(-1) \odot x = x^{-1} = 1/x$ , exactly what Part 4 says must happen.

3.3Going deeper

Vectors need not be tuples. In the last two examples the "vectors" are matrices with fixed 1s and positive real numbers. What matters is the pair (set, operations).
The zero vector moves. It was $(0, \dots, 0)$ in $\mathbb{R}^n$ , $(0, 1)$ in $\{(x, 1)\}$ , $(-1, 0)$ in the shifted plane, $I_2$ in the matrix example and $1$ in the positive reals.
Many are just $\mathbb{R}^n$ in disguise (extra). A relabelling that respects both operations (like $\varphi$ ) shows the strange spaces behave exactly like ordinary ones. For positive reals the relabelling is $x \mapsto \ln x$ .

3.4Exam tips and common mistakes

Exam tips

Find the zero vector first. If none exists, you are done: not a vector space.
Then find the inverse using the zero vector you found, not (0, 0).

Common mistakes

Using the usual zero when the operations are unusual.
Declaring "not a vector space" because (0, 0) is missing from the set. Check the actual zero vector.

Practice questions (graded-assignment style)

Q1The shifted zero

Numerical Answer Type (NAT).

In the shifted plane ( $(x_1, y_1) + (x_2, y_2) = (x_1 + x_2 + 1,\ y_1 + y_2)$ ), what is the first entry of the zero vector?

Solution

Idea

Solve $(x, y) + (m, n) = (x, y)$ .

$(x + m + 1,\ y + n) = (x, y)$ gives $m = -1$ , $n = 0$ .
So $0_V = (-1, 0)$ and the first entry is $-1$ .
Answer: -1

Q2An inverse in the shifted plane

Numerical Answer Type (NAT).

In the same space, what is the first entry of the additive inverse of $(3, 2)$ ?

Solution

Idea

The inverse of $(x, y)$ is $(-x - 2,\ -y)$ .

$(3, 2) + (a, b) = (3 + a + 1,\ 2 + b) = (-1, 0)$ gives $a = -5$ , $b = -2$ .
The first entry is $-5$ .
Answer: -5

Q3Zero of the positive reals

Multiple Choice Question (MCQ): choose the one correct option.

For $V = (0, \infty)$ with $x \oplus y = xy$ , $c \odot x = x^c$ , the zero vector is:

−1

There is none

Solution

Idea

Solve $x \oplus m = x$ .

$x \oplus m = xm = x$ , so $m = 1$ . ( $0$ is not even in $V$ .)
Answer: option 2.

Q4Inverse of 5

Numerical Answer Type (NAT).

In the same space, what is the additive inverse of 5? (Give a decimal.)

Solution

Idea

$5 \oplus x\prime = 1$ .

$5x\prime = 1$ , so $x\prime = \tfrac15 = 0.2$ .
Answer: 0.2

Q5Vector spaces under the usual operations

Multiple Select Question (MSQ): select all that apply.

Which of these are vector spaces under the usual addition and scalar multiplication?

\mathbb{R}^4

All

2 \times 3

real matrices

\{x \in \mathbb{R}^3 : Ax = 0\}

\{(x, y) : y = 1\}

Solution

Idea

Is each set closed under + and scalars, and does it contain 0?

Options 1, 2, 3 are vector spaces (Examples 7, 8, 9).
Option 4: $(x, 1) + (y, 1) = (x + y, 2)$ leaves the set, so with the usual operations it is not even closed.
Answer: options 1, 2 and 3.

Q6An inverse in the matrix example

Numerical Answer Type (NAT).

In the space of matrices $\begin{bmatrix}1&a\\b&1\end{bmatrix}$ (Example 13), take $u = \begin{bmatrix}1&3\\-2&1\end{bmatrix}$ . If its additive inverse is $\begin{bmatrix}1&a\prime\\b\prime&1\end{bmatrix}$ , find $a\prime + b\prime$ .

Solution

Idea

Off-diagonal entries add to 0.

$a\prime = -3$ and $b\prime = 2$ .
$a\prime + b\prime = -1$ .
Answer: -1

Revision overview

Standard spacesℝⁿ, M_{m×n}(ℝ)
solutions of Ax = 0

Strange operationsCheck all nine axioms
Or find one that fails

Zero vector(0,1) in {(x,1)}
(−1,0) shifted plane
1 in positive reals

Is it a vector space?Week 3 · Part 3

FailsStrange addition: axiom 3
Ignoring entry: axiom 5

Passes{(x,1)}, shifted plane,
matrices with 1s, (0,∞)

IdeaOften ℝⁿ in disguise

Revision: verifying or breaking the axioms in concrete examples

Next · FoundationsProperties of Vector Spaces

Foundations

Properties of Vector Spaces

Facts that follow from the axioms alone, and so hold in every vector space.

Everything below follows from the nine axioms alone, so it holds in every vector space, including the strange ones from Part 3.

4.1The cancellation law

Theorem: cancellation law

Let $V$ be a vector space and $v_1, v_2, v_3 \in V$ . If $v_1 + v_3 = v_2 + v_3$ , then $v_1 = v_2$ .

Example 15Proof of the cancellation law

Prove: $v_1 + v_3 = v_2 + v_3 \Rightarrow v_1 = v_2$ .

Solution

$v_3$ has an additive inverse $v_3\prime$ with $v_3 + v_3\prime = 0_V$ (axiom 6).
Add $v_3\prime$ to both sides of $v_1 + v_3 = v_2 + v_3$ : $(v_1 + v_3) + v_3\prime = (v_2 + v_3) + v_3\prime$ .
Associativity: $v_1 + (v_3 + v_3\prime) = v_2 + (v_3 + v_3\prime)$ , i.e. $v_1 + 0_V = v_2 + 0_V$ .
The zero vector leaves vectors unchanged, so $v_1 = v_2$ . $\blacksquare$ (Similarly, $v_3 + v_1 = v_3 + v_2$ gives $v_1 = v_2$ , using commutativity.)

4.2Uniqueness of zero and inverses

Corollary

The zero vector of $V$ is unique.
Each $u \in V$ has exactly one additive inverse, written $-u$ .

Proof of (i). If $0_V$ and $0_V\prime$ both act as zero, then $0_V\prime + 0_V = 0_V\prime$ (since $0_V$ is a zero) and $0_V\prime + 0_V = 0_V$ (since $0_V\prime$ is a zero). So $0_V\prime = 0_V$ .
Proof of (ii). If $u\prime$ and $w$ are both inverses of $u$ , then $u + u\prime = 0_V = u + w$ , and cancelling $u$ gives $u\prime = w$ .

4.3Scalars meeting zero

Corollary

For every $u \in V$ and $c \in \mathbb{R}$ :

$0 \cdot u = 0_V$
$c \cdot 0_V = 0_V$
$(-c)\cdot u = -(c\cdot u) = c\cdot(-u)$

(i) Since $0 = 0 + 0$ in $\mathbb{R}$ : $0\cdot u = (0 + 0)\cdot u = 0\cdot u + 0\cdot u$ , so $0\cdot u + 0\cdot u = 0\cdot u + 0_V$ , and cancellation gives $0\cdot u = 0_V$ .
(ii) $0_V = 0_V + 0_V$ , so $c\cdot 0_V = c\cdot 0_V + c\cdot 0_V$ ; cancelling gives $c\cdot 0_V = 0_V$ .
(iii) $0_V = 0\cdot u = (c + (-c))\cdot u = c\cdot u + (-c)\cdot u$ , so $(-c)\cdot u$ is the (unique) inverse of $c\cdot u$ . Likewise $c\cdot u + c\cdot(-u) = c\cdot 0_V = 0_V$ .

A handy consequence

In any vector space $0_V = 0\cdot u$ and $-u = (-1)\cdot u$ . You can always compute the zero vector and inverses from scalar multiplication. In the shifted plane: $0\cdot(x, y) = (0 + 0 - 1,\ 0) = (-1, 0)$ and $(-1)\cdot(x, y) = (-x - 1 - 1,\ -y) = (-x - 2,\ -y)$ , matching the answers found before.

4.4Going deeper

Why these results matter. They let you prove facts once for all vector spaces and then use them in the weird examples without recomputing.
Test for the zero vector. If you computed a "zero vector" $z$ in an example, check that $0\cdot u = z$ for some $u$ . If it fails, the example is not a vector space or you made an error (extra).

4.5Exam tips and common mistakes

Exam tips

In a proof, name the axiom behind every line (associativity, inverse, ...).
To prove uniqueness, assume two objects work and show they are equal.

Common mistakes

Using properties of real numbers (like dividing by a vector) that do not exist for vectors.
Confusing the scalar $0$ with the zero vector $0_V$ .

Practice questions (graded-assignment style)

Q1Which law?

Multiple Choice Question (MCQ): choose the one correct option.

From $v_1 + v_3 = v_2 + v_3$ we may conclude $v_1 = v_2$ . This is called:

The distributive law

The cancellation law

The unity axiom

Closure

Solution

Idea

Add the inverse of v₃ to both sides.

Adding $v_3\prime$ to both sides and using associativity and the zero vector gives $v_1 = v_2$ : the cancellation law.
Answer: option 2.

Q2Consequences of the axioms

Multiple Select Question (MSQ): select all that apply.

In every vector space, which statements hold for all $u$ and all scalars $c$ ?

0\cdot u = 0_V

c\cdot 0_V = 0_V

(-1)\cdot u = -u

0\cdot u = u

Solution

Idea

Recall Corollary 3.3.3.

Options 1, 2, 3 are proved in Part 4.
Option 4 is false: $0\cdot u = 0_V$ , which equals $u$ only when $u = 0_V$ .
Answer: options 1, 2 and 3.

Q3The zero from scalar multiplication

Numerical Answer Type (NAT).

In the positive reals with $x \oplus y = xy$ , $c \odot x = x^c$ , compute $0 \odot 7$ .

Solution

Idea

$0\cdot u$ is always the zero vector.

$0 \odot 7 = 7^0 = 1$ , which is $0_V$ of this space, as Corollary (i) predicts.
Answer: 1

Q4Inverse as (−1)·u

Numerical Answer Type (NAT).

In the shifted plane, compute the first entry of $(-1)\cdot(4, 5)$ , where $c\cdot(x, y) = (cx + c - 1,\ cy)$ .

Solution

Idea

$(-1)\cdot u = -u$ .

$(-1)\cdot(4, 5) = (-4 - 1 - 1,\ -5) = (-6, -5)$ .
This matches $-u = (-x - 2,\ -y) = (-6, -5)$ .
Answer: -6

Q5If cu = 0

Multiple Choice Question (MCQ): choose the one correct option.

If $c\cdot u = 0_V$ and $c \neq 0$ , then:

u = 0_V

u

can be any vector

c = 1

No such u exists

Solution

Idea

Multiply by $1/c$ .

$u = \tfrac1c(c\cdot u) = \tfrac1c\cdot 0_V = 0_V$ .
Answer: option 1.

Revision overview

Cancellationv₁ + v₃ = v₂ + v₃
⟹ v₁ = v₂

UniquenessOne zero vector
One inverse per vector

Scalars and zero0·u = 0_V
c·0_V = 0_V
(−c)u = −(cu)

Properties of vector spacesWeek 3 · Part 4

Handy0_V = 0·u
−u = (−1)·u

Proofs−(−u) = u
cu = 0 ⟹ c = 0 or u = 0

ConsequenceA non-zero vector space
is infinite

Revision: what follows from the nine axioms alone

Next · SubspacesSubspaces

Subspaces

Vector spaces living inside vector spaces, and the two-step test that recognises them.

5.1The definition and the two-step test

Definition: subspace

A non-empty subset $W$ of a vector space $V$ is a subspace of $V$ if $W$ is itself a vector space under the same addition and scalar multiplication as $V$ .

Theorem: the subspace test

A non-empty subset $W$ of $V$ is a subspace if and only if

$w_1, w_2 \in W \Rightarrow w_1 + w_2 \in W$ (closed under addition), and
$c \in \mathbb{R},\ w_1 \in W \Rightarrow c\cdot w_1 \in W$ (closed under scalar multiplication).

Why only two? Axioms 3, 4, 7, 8, 9 are equations that already hold for all vectors of $V$ , so they hold for vectors of $W$ automatically.
Zero and inverses come free. Pick any $w \in W$ : $0_V = 0\cdot w \in W$ and $-w = (-1)\cdot w \in W$ by Part 4.
Trivial subspaces. Every vector space $V$ has two: $V$ itself and $\{0_V\}$ (since $0_V + 0_V = 0_V$ and $c\cdot 0_V = 0_V$ ). A subspace $W \subsetneq V$ is called proper.

The quick zero test

Every subspace contains $0_V$ . So if $0_V \notin W$ , stop: $W$ is not a subspace. The converse is false: the parabola below contains $(0, 0)$ and still fails.

Figure 4. The subspace test as four checks

5.2Examples that fail

Figure 5. Adding or scaling points of y = x² throws you off the curve

5.3Lines, systems and half-spaces: three proofs

Example 24A line is a subspace iff it passes through the origin

Prove that a line $L = \{(x, y) : ax + by = d\}$ (with $(a, b) \neq (0, 0)$ ) in $\mathbb{R}^2$ is a subspace if and only if it passes through the origin.

Solution

"Passes through the origin" means $(0, 0) \in L$ : $a\cdot 0 + b\cdot 0 = d$ , i.e. $d = 0$ . So we show: $L$ is a subspace $\iff d = 0$ .
$(\Rightarrow)$ A line is non-empty; pick $w \in L$ . Closure under scalars gives $0\cdot w = (0, 0) \in L$ . Substituting into the equation: $0 = d$ .
$(\Leftarrow)$ Let $d = 0$ . Take $(x_1, y_1), (x_2, y_2) \in L$ : $ax_1 + by_1 = 0$ and $ax_2 + by_2 = 0$ . Their sum satisfies $a(x_1 + x_2) + b(y_1 + y_2) = 0 + 0 = 0$ .
For $c \in \mathbb{R}$ : $a(cx_1) + b(cy_1) = c(ax_1 + by_1) = 0$ .
$L$ is non-empty and closed under both, so it is a subspace. $\blacksquare$

Figure 6. Same direction, but only the line through O is a subspace

Example 25Solutions of $A\mathbf{x} = \mathbf{b}$

Let $W$ be the set of solutions of $A\mathbf{x} = \mathbf{b}$ in $\mathbb{R}^n$ , where $A$ is $m \times n$ . Show $W$ is a subspace if and only if $\mathbf{b} = \mathbf{0}$ .

Solution

$(\Leftarrow)$ If $\mathbf{b} = \mathbf{0}$ : $A\cdot 0 = 0$ , so $0 \in W$ (non-empty). For $u, v \in W$ and $c \in \mathbb{R}$ : $A(u + v) = Au + Av = 0$ and $A(cu) = c(Au) = 0$ . By the subspace test, $W$ is a subspace.
$(\Rightarrow)$ If $W$ is a subspace, it is non-empty and contains the zero vector (take $w \in W$ , then $0\cdot w = 0 \in W$ ). So $x = 0$ is a solution: $\mathbf{b} = A\cdot 0 = \mathbf{0}$ .
So $W$ is a subspace exactly when $\mathbf{b} = \mathbf{0}$ . (If $\mathbf{b} \neq \mathbf{0}$ , then either the system is inconsistent and $W$ is empty, or $W$ is non-empty but misses $0$ ; both rule out a subspace.) $\blacksquare$

Example 26A half-space is not a subspace

Show that $W = \{(x, y, z) : x \geq z\}$ is not a subspace of $\mathbb{R}^3$ .

Solution

Zero test. $0 \geq 0$ is true, so $0 \in W$ : the test does not decide it.
Addition. If $x_1 \geq z_1$ and $x_2 \geq z_2$ , adding the inequalities gives $x_1 + x_2 \geq z_1 + z_2$ : $W$ is closed under addition.
Scalars. Multiplying an inequality by a negative number reverses it, so try a negative scalar. $(1, 0, 0) \in W$ since $1 \geq 0$ . But $(-1)\cdot(1, 0, 0) = (-1, 0, 0)$ , and $-1 \geq 0$ is false.
$W$ is not closed under scalar multiplication, so it is not a subspace. (Geometrically a half-space: flipping by $-1$ sends points to the other side of the plane $x = z$ .)

5.4Going deeper

Intersections are subspaces. If $W_1, W_2$ are subspaces of $V$ , then $0_V$ lies in both, and if $u, v$ lie in both then so do $u + v$ and $cu$ . Example: in $\mathbb{R}^3$ , the $xy$ -plane $\cap$ the $xz$ -plane $=$ the $x$ -axis.
Unions usually are not (extra). The $x$ -axis $\cup$ the $y$ -axis contains $(1, 0)$ and $(0, 1)$ but not their sum $(1, 1)$ .
Subspaces of $\mathbb{R}^3$ (extra). They are exactly: $\{0\}$ , lines through the origin, planes through the origin, and $\mathbb{R}^3$ .

Figure 7. Subspaces of ℝ³ nest inside each other

5.5Exam tips and common mistakes

Exam tips

Run the zero test first; it kills most "plane x + y + z = 1" type sets in one line.
If the zero test passes, try a negative scalar and a sum of two simple vectors as counterexamples.
To prove a subspace, take general elements and show both closures.

Common mistakes

Concluding "subspace" from 0 ∈ W alone.
Testing closure with one example. A proof needs general elements.
Forgetting negative scalars for inequalities.

Practice questions (graded-assignment style)

Q1Which are subspaces?

Multiple Select Question (MSQ): select all that apply.

Which of the following are subspaces of $\mathbb{R}^3$ ?

\{(0, y, z)\}

\{x + y + z = 1\}

\{x + y + z = 0\}

\{x \geq z\}

Solution

Idea

Zero test, then closure.

Options 1 and 3 pass both closures.
Option 2 misses the origin. Option 4 fails scalar closure with $-1$ .
Answer: options 1 and 3.

Q2Which d?

Numerical Answer Type (NAT).

For which value of $k$ is $\{(x, y) : x + 2y = k\}$ a subspace of $\mathbb{R}^2$ ?

Solution

Idea

A line is a subspace iff it passes through the origin.

$(0, 0)$ must satisfy $0 + 0 = k$ , so $k = 0$ .
Answer: 0

Q3Unions and intersections

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about subspaces of a vector space $V$ .

The intersection of two subspaces is a subspace.

The union of two subspaces is always a subspace.

Every subspace contains

0_V

Every subset containing

0_V

is a subspace.

Solution

Idea

Try the x-axis and y-axis for unions.

Statements 1 and 3 are true.
Union: $(1, 0)$ and $(0, 1)$ are in the union of the axes, but $(1, 1)$ is not. Statement 4: the parabola contains 0 and is not a subspace.
Answer: options 1 and 3.

Q4Invertible matrices

Multiple Choice Question (MCQ): choose the one correct option.

The set of invertible $2 \times 2$ matrices is not a subspace of $M_{2\times 2}(\mathbb{R})$ because:

It is empty

The zero matrix is not invertible

It is not closed under scalar multiplication by 2

It has no identity

Solution

Idea

What is the zero vector of M₂ₓ₂?

The zero vector of $M_{2\times 2}$ is the zero matrix, which has determinant 0, so it is not in the set.
Answer: option 2.

Q5A counterexample scalar

Numerical Answer Type (NAT).

For $W = \{(x, y, z) : x \geq z\}$ and $(1, 0, 0) \in W$ , find the largest integer $c$ for which $c\cdot(1, 0, 0) \notin W$ .

Solution

Idea

$c\cdot(1, 0, 0) = (c, 0, 0)$ is in $W$ iff $c \geq 0$ .

For $c \geq 0$ the point is in $W$ ; for $c < 0$ it is not.
The largest negative integer is $-1$ .
Answer: -1

Revision overview

TestNon-empty
w₁ + w₂ ∈ W
c·w₁ ∈ W

Quick check0_V ∈ W necessary,
not sufficient

TrivialV itself and {0_V}

SubspacesWeek 3 · Part 5

PassesLines and planes through O
symmetric matrices
solutions of Ax = 0

Failsy = x², x + y + z = 1
x ≥ z, invertible matrices
Ax = b with b ≠ 0

CombineIntersections stay subspaces

Revision: deciding whether a subset is a subspace

Next · Building blocksLinear Combinations

Building blocks

Linear Combinations

Building new vectors from old ones with scalars and addition.

A non-zero vector space has infinitely many vectors. Chapter 4 looks for a small, finite set of vectors, a basis, that describes the whole space. The building blocks are linear combinations, linear dependence and span.

6.1Definition

Definition: linear combination

Let $V$ be a vector space and $v_1, \dots, v_n \in V$ . Then

\sum_{i=1}^{n} \alpha_i v_i = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n

is a linear combination of $v_1, \dots, v_n$ with coefficients $\alpha_1, \dots, \alpha_n \in \mathbb{R}$ . Since $V$ is closed under addition and scalar multiplication, a linear combination is again a vector in $V$ .

6.2A picture in the plane

Example 27The parallelogram law

Show that $(4, 5)$ is a linear combination of $(1, 2)$ and $(2, 1)$ .

Solution

Scale $(1, 2)$ by 2 to get $(2, 4)$ .
The parallelogram with sides $(2, 4)$ and $(2, 1)$ has diagonal $(4, 5)$ ; algebraically $(2, 4) + (2, 1) = (4, 5)$ .
So $(4, 5) = 2(1, 2) + 1\cdot(2, 1)$ : a combination with $\alpha_1 = 2$ , $\alpha_2 = 1$ .
Any one of the three can be written with the other two: $(2, 1) = (4, 5) - 2(1, 2)$ and $(1, 2) = \tfrac12(4, 5) - \tfrac12(2, 1)$ .
Moving everything to one side: $2(1, 2) + (2, 1) - (4, 5) = (0, 0)$ , the zero vector written with non-zero coefficients. This is what "linearly dependent" will mean (Part 7).

Figure 8. 2(1, 2) + (2, 1) = (4, 5)

6.3A picture in space

Example 28Three vectors in one plane, and one off it

Show that $(3, 7, 2) = 2(0, 2, 1) + \tfrac32(2, 2, 0)$ , that all three vectors lie on the plane $2x - 2y + 4z = 0$ , and that $(1, 1, 1)$ is not a combination of $(0, 2, 1)$ and $(2, 2, 0)$ .

Solution

$2(0, 2, 1) = (0, 4, 2)$ , $\tfrac32(2, 2, 0) = (3, 3, 0)$ , and $(0, 4, 2) + (3, 3, 0) = (3, 7, 2)$ . $\checkmark$
Again each vector is a combination of the other two, e.g. $(2, 2, 0) = \tfrac23(3, 7, 2) - \tfrac43(0, 2, 1)$ .
Plane check: $(0, 2, 1)$ : $0 - 4 + 4 = 0$ . $(2, 2, 0)$ : $4 - 4 + 0 = 0$ . $(3, 7, 2)$ : $6 - 14 + 8 = 0$ . $\checkmark$
$(1, 1, 1)$ is off the plane: $2 - 2 + 4 = 4 \neq 0$ . Suppose $(1, 1, 1) = a(0, 2, 1) + b(2, 2, 0) = (2b,\ 2a + 2b,\ a)$ . Then $2b = 1$ , $2a + 2b = 1$ , $a = 1$ : so $b = \tfrac12$ , $a = 1$ , but $2a + 2b = 3 \neq 1$ . Contradiction, so it is not a combination.
Consequently $\alpha(1, 1, 1) + \beta(0, 2, 1) + \gamma(2, 2, 0) = 0$ only when $\alpha = \beta = \gamma = 0$ . (If $\alpha \neq 0$ we could solve for $(1, 1, 1)$ , impossible; so $\alpha = 0$ , and then $2\gamma = 0$ forces $\gamma = 0$ and $\beta = 0$ .)

Figure 9. Three vectors on one plane through O, and (1, 1, 1) sticking out of it

6.4How to test "is v a combination?"

Write a general combination, match it with $v$ entry by entry, and solve the resulting linear system for the coefficients. A solution means yes; an inconsistent system means no.

Example 29Is it a combination?

Is $(3, 5, 1)$ a linear combination of $(1, 1, 0)$ and $(0, 1, 1)$ ? What about $(3, 4, 1)$ ?

Solution

A general combination is $a(1, 1, 0) + b(0, 1, 1) = (a,\ a + b,\ b)$ .
Match with $(3, 5, 1)$ : $a = 3$ , $a + b = 5$ , $b = 1$ . The first and third give $a = 3$ , $b = 1$ , but then $a + b = 4 \neq 5$ . Inconsistent: no.
Match with $(3, 4, 1)$ : $a = 3$ , $a + b = 4$ , $b = 1$ : consistent. Yes: $(3, 4, 1) = 3(1, 1, 0) + 1\cdot(0, 1, 1)$ .
Geometrically every combination $(a, a + b, b)$ has middle entry = first + last, i.e. lies on the plane $x - y + z = 0$ . $(3, 5, 1)$ : $3 - 5 + 1 = -1$ (off the plane). $(3, 4, 1)$ : $3 - 4 + 1 = 0$ (on it).

6.5Going deeper

Combinations of two non-parallel vectors fill a plane through the origin; combinations of one non-zero vector fill a line (extra, made precise in Part 9).
Coefficients can be anything. They may be zero, negative or fractions; only the vectors matter, not whether the coefficients are "nice".
Everything is a combination of itself. $v = 1\cdot v$ , and $0_V = 0\cdot v$ is a combination of any vectors (all coefficients 0).

6.6Exam tips and common mistakes

Exam tips

Set up the system by comparing entries, then solve; do not guess coefficients.
For "must it be a combination?" questions look for a counterexample with zero or parallel vectors.

Common mistakes

Checking only two of the three equations.
Dividing by a coefficient that may be zero when solving for a vector.

Practice questions (graded-assignment style)

Q1Find the unknown vector

Numerical Answer Type (NAT).

If $(9, 3, 1)$ is a linear combination of $(1, 1, 1)$ , $(1, -1, 1)$ and $(x, y, z)$ as $2(1, 1, 1) + 3(1, -1, 1) + 4(x, y, z) = (9, 3, 1)$ , find $x + y + 2z$ .

Solution

Idea

Isolate 4(x, y, z).

$2(1,1,1) + 3(1,-1,1) = (2,2,2) + (3,-3,3) = (5, -1, 5)$ .
$4(x, y, z) = (9, 3, 1) - (5, -1, 5) = (4, 4, -4)$ , so $(x, y, z) = (1, 1, -1)$ .
Check: $2 + 3 + 4 = 9$ , $2 - 3 + 4 = 3$ , $2 + 3 - 4 = 1$ . $x + y + 2z = 1 + 1 - 2 = 0$ .
Answer: 0

Q2Must it be a combination?

Multiple Select Question (MSQ): select all that apply.

Let $v_1 = av_2 + bv_3 + cv_4$ in a vector space $V$ . Which statements must be true?

v_2

is a linear combination of

v_i,\ i = 1, 3, 4

v_3

is a linear combination of

v_i,\ i = 1, 4

v_2

is a linear combination of

v_2

v_4

is a linear combination of

v_4

Solution

Idea

Any vector is a combination of itself. For the others, try counterexamples.

Options 3 and 4: $v = 1\cdot v$ . Always true.
Option 1 needs dividing by $a$ , which may be 0. Counterexample in $\mathbb{R}^2$ : $v_2 = (0, 1)$ , $v_3 = (1, 0)$ , $v_4 = (0, 0)$ , $v_1 = (1, 0) = 0\cdot v_2 + 1\cdot v_3 + 0\cdot v_4$ ; combinations of $v_1, v_3, v_4$ are $(\alpha, 0)$ , never $(0, 1)$ .
Option 2 fails similarly: $v_2 = (1, 0)$ , $v_3 = (0, 1)$ , $v_4 = (0, 0)$ , $v_1 = (1, 0)$ ; combinations of $v_1, v_4$ are $(\alpha, 0)$ , never $(0, 1)$ .
Answer: options 3 and 4.

Q3Which is a combination?

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about $u = (1, 1, 0)$ and $w = (0, 1, 1)$ .

(3, 5, 1)

is a linear combination of

u, w

(3, 4, 1)

is a linear combination of

u, w

(3, 4, 1) = 3u + w

Every combination

(x, y, z)

u, w

satisfies

x - y + z = 0

Solution

Idea

A combination is (a, a + b, b).

$(3, 5, 1)$ : $a = 3, b = 1$ forces $a + b = 4 \neq 5$ . Not a combination.
$(3, 4, 1) = 3(1,1,0) + (0,1,1)$ and $a - (a + b) + b = 0$ always.
Answer: options 2, 3 and 4.

Q4Find the coefficient

Numerical Answer Type (NAT).

$(5, 7) = a(1, 1) + b(1, 2)$ . Find $a$ .

Solution

Idea

a + b = 5 and a + 2b = 7.

Subtract: $b = 2$ , then $a = 3$ .
Check: $3(1,1) + 2(1,2) = (5, 7)$ .
Answer: 3

Q5Rewriting the dependence

Multiple Choice Question (MCQ): choose the one correct option.

From $2(1, 2) + (2, 1) - (4, 5) = (0, 0)$ which statement follows?

(1, 2) = \tfrac12(4, 5) - \tfrac12(2, 1)

(1, 2) = (4, 5) + (2, 1)

(1, 2) = 2(4, 5)

None of these

Solution

Idea

Solve for (1, 2).

$2(1, 2) = (4, 5) - (2, 1)$ , so $(1, 2) = \tfrac12(4, 5) - \tfrac12(2, 1)$ .
Answer: option 1.

Revision overview

Combinationα₁v₁ + … + αₙvₙ
Coefficients real

PictureParallelogram law in ℝ²
Plane through O in ℝ³

Stays insideA combination of vectors
of V is in V

Linear combinationsWeek 3 · Part 6

TestMatch entries
Solve for coefficients

Example(4,5) = 2(1,2) + (2,1)

NextZero as a combination:
dependence

Revision: building vectors from other vectors

Next · Building blocksLinear Dependence and Independence

Building blocks

Linear Dependence and Independence

When a set contains redundancy, and when it does not.

7.1Linear dependence

Definition: linearly dependent

A set of vectors $\{v_1, \dots, v_n\}$ is linearly dependent if there exist scalars $\alpha_1, \dots, \alpha_n \in \mathbb{R}$ , not all zero, such that

\alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n = 0

In words: the zero vector can be built from the vectors with at least one non-zero coefficient. Geometrically, any two vectors on the same line, or any three vectors on the same plane through the origin, are dependent. From Part 6, $\{(4, 5), (2, 1), (1, 2)\}$ and $\{(3, 7, 2), (0, 2, 1), (2, 2, 0)\}$ are both dependent.

Example 30Showing a set is dependent

Show that $\{(1, 2, 4),\ (2, -1, 2),\ (5, 0, 8)\}$ is linearly dependent.

Solution

Set up. $a(1, 2, 4) + b(2, -1, 2) + c(5, 0, 8) = (0, 0, 0)$ gives $a + 2b + 5c = 0$ , $2a - b = 0$ , $4a + 2b + 8c = 0$ .
Solve. From the second, $b = 2a$ . Then the first is $a + 4a + 5c = 0$ , so $c = -a$ . The third becomes $4a + 4a - 8a = 0$ , true for every $a$ . So $(a, b, c) = (a,\ 2a,\ -a)$ , with $a$ free.
Non-zero solution. $a = -1$ gives $b = -2$ , $c = 1$ : $-1(1, 2, 4) - 2(2, -1, 2) + 1(5, 0, 8) = (-1 - 4 + 5,\ -2 + 2 + 0,\ -4 - 4 + 8) = (0, 0, 0)$ .
Non-zero coefficients produce 0, so the set is linearly dependent.
Adding $(1, 1, 1)$ keeps it dependent: $-1(1,2,4) - 2(2,-1,2) + 1(5,0,8) + 0\cdot(1,1,1) = 0$ . Every superset of a dependent set is dependent (give the extra vectors coefficient 0).

Remark: a set with the zero vector is dependent. If

v_i = 0

, take

\alpha_i = 1

and every other coefficient 0: then

\sum \alpha_j v_j = 1\cdot 0 = 0

with a non-zero coefficient.

7.2Linear independence

Definition: linearly independent

Vectors $v_1, \dots, v_n$ are linearly independent if they are not linearly dependent, that is,

\alpha_1 v_1 + \cdots + \alpha_n v_n = 0 \;\Longrightarrow\; \alpha_1 = \alpha_2 = \cdots = \alpha_n = 0

The only way to get the zero vector is the trivial way, with all coefficients zero.

Example 31Showing a pair is independent

Is $\{(1, 1),\ (1, -1)\}$ linearly independent?

Solution

Suppose $a(1, 1) + b(1, -1) = (0, 0)$ : $a + b = 0$ and $a - b = 0$ .
Adding: $2a = 0$ , so $a = 0$ , and then $b = 0$ .
Only the trivial solution, so the set is linearly independent. (Equivalently, $(1, -1)$ is not a multiple of $(1, 1)$ .)

Figure 10. Two vectors: one a multiple of the other (dependent) versus not (independent)

7.3Two vectors, three vectors

Two vectors are dependent exactly when one is a scalar multiple of the other. For non-zero $v_1, v_2$ with $\alpha_1 v_1 + \alpha_2 v_2 = 0$ , $\alpha_1, \alpha_2$ not both zero: if exactly one were zero, say $\alpha_1 = 0$ , then $\alpha_2 v_2 = 0$ would force $v_2 = 0$ , a contradiction. So both are non-zero and $v_1 = -\tfrac{\alpha_2}{\alpha_1} v_2$ .
Three non-zero vectors are independent iff none is a combination of the other two. Geometrically, three vectors in $\mathbb{R}^3$ are independent exactly when they do not lie on one plane through the origin. Compare $(3, 7, 2), (0, 2, 1), (2, 2, 0)$ (dependent) with $(1, 1, 1), (0, 2, 1), (2, 2, 0)$ (independent).

Example 32Three vectors in R³

Show $\{(1, 1, 2),\ (-1, 0, 1),\ (2, 1, 2)\}$ is linearly independent.

Solution

Suppose $a(1, 1, 2) + b(-1, 0, 1) + c(2, 1, 2) = 0$ : $a - b + 2c = 0$ , $a + c = 0$ , $2a + b + 2c = 0$ .
From the second, $a = -c$ . The first becomes $-c - b + 2c = 0$ , so $b = c$ .
The third becomes $-2c + c + 2c = c = 0$ . Therefore $c = 0$ , $b = 0$ , $a = 0$ : only the trivial solution, so the set is independent.

7.4Going deeper

Supersets and subsets. A superset of a dependent set is dependent; a subset of an independent set is independent. (A non-trivial combination of a subset would be a non-trivial combination of the whole set.)
Removing a vector can break dependence. $\{(1, 0, 0), (2, 0, 0)\}$ is dependent, but removing $(2, 0, 0)$ leaves the independent set $\{(1, 0, 0)\}$ .
Union and intersection (extra). The union of two independent sets may be dependent ( $\{(1,0)\} \cup \{(2,0)\}$ ) or independent ( $\{(1,0)\} \cup \{(0,1)\}$ ). The intersection is always independent, being a subset.

7.5Exam tips and common mistakes

Exam tips

Two vectors: just check whether one is a multiple of the other.
To show dependence, one non-trivial combination is enough, so quote it explicitly.
To show independence, solve the system and show only the trivial solution.

Common mistakes

Writing "0·v₁ + 0·v₂ = 0, so independent". The zero combination always works; it proves nothing.
Forgetting that a set containing the zero vector is dependent.
Thinking removing a vector from a dependent set must keep it dependent.

Practice questions (graded-assignment style)

Q1A dependent set S

Multiple Select Question (MSQ): select all that apply.

Let $S$ be a linearly dependent subset of $\mathbb{R}^3$ . Which statements are true?

S \cup \{(1, 0, 0)\}

must be linearly dependent

S \cup \{v\}

must be linearly dependent for any

v \in \mathbb{R}^3

S \setminus \{v\}

must be linearly dependent for any

v \in S

There may exist some

v \in S

such that

S \setminus \{v\}

is still dependent

Solution

Idea

Reuse the same coefficients and give the new vector coefficient 0.

Options 1 and 2: the old non-trivial combination plus $0\cdot v$ shows dependence.
Option 3 fails: $S = \{(1,0,0), (2,0,0)\}$ is dependent, but removing $(2,0,0)$ leaves $\{(1,0,0)\}$ , independent.
Option 4: $S = \{(1,0,0),(2,0,0),(0,1,0)\}$ ; removing $(0,1,0)$ leaves a dependent set. True.
Answer: options 1, 2 and 4.

Q2Which equation proves what?

Multiple Select Question (MSQ): select all that apply.

Choose the correct statement(s).

0(0, 0) + 0(1, 1) = (0, 0)

implies

\{(0, 0), (1, 1)\}

is independent

2(1, 0) + 2(0, 1) = (2, 2)

implies

\{(1, 0), (0, 1), (2, 2)\}

is dependent

2(1, 0) + 2(0, 1) = (2, 2)

implies

\{(1, 0), (2, 2)\}

is dependent

Solution

Idea

Make the zero vector appear, and check that every vector in the equation is in the set.

Option 1: the all-zero combination always works. The set contains $(0,0)$ so it is dependent.
Option 2: $2(1,0) + 2(0,1) - 1\cdot(2,2) = (0,0)$ with non-zero coefficients: dependent. True.
Option 3 uses $(0, 1)$ , which is not in that set. Directly, $(2, 2)$ is not a multiple of $(1, 0)$ ( $\det = 2 \neq 0$ ): independent.
Answer: option 2.

Q3Unions and intersections

Multiple Select Question (MSQ): select all that apply.

Choose the correct options.

The union of two distinct independent sets may be dependent.

The union of two distinct independent sets must be dependent.

A non-empty intersection of two independent sets must be independent.

A non-empty intersection of two independent sets must be dependent.

Solution

Idea

Find examples; recall subsets of independent sets.

$\{(1,0)\}$ and $\{(2,0)\}$ are independent but their union is dependent (option 1 true).
$\{(1,0)\}$ and $\{(0,1)\}$ have an independent union (option 2 false).
An intersection is a subset of an independent set, so it is independent (option 3 true, option 4 false).
Answer: options 1 and 3.

Q4Make it dependent

Numerical Answer Type (NAT).

For which $k$ is $\{(1, 2), (2, k)\}$ linearly dependent?

Solution

Idea

Two vectors are dependent iff one is a multiple of the other.

$(2, k) = 2(1, 2) = (2, 4)$ , so $k = 4$ .
Check with the determinant: $1\cdot k - 2\cdot 2 = 0$ gives $k = 4$ .
Answer: 4

Q5Three vectors in the plane

Multiple Choice Question (MCQ): choose the one correct option.

The set $\{(1, 0), (0, 1), (1, 1)\}$ in $\mathbb{R}^2$ is:

Linearly independent

Linearly dependent

Empty

A subspace of R³

Solution

Idea

Find a non-trivial combination giving 0.

$1(1,0) + 1(0,1) - 1(1,1) = (0,0)$ with non-zero coefficients, so it is dependent.
(Also: three vectors in $\mathbb{R}^2$ are always dependent; see Part 8.)
Answer: option 2.

Revision overview

DependentΣαᵢvᵢ = 0 with some αᵢ ≠ 0
Contains 0 ⟹ dependent

IndependentΣαᵢvᵢ = 0 ⟹ all αᵢ = 0

Two vectorsDependent iff
one is a multiple of the other

Dependence and independenceWeek 3 · Part 7

SupersetsSuperset of dependent
is dependent

SubsetsSubset of independent
is independent

Geometry3 vectors on a plane
through O: dependent

Revision: telling dependent sets from independent ones

Next · Building blocksChecking Independence With Matrices

Building blocks

Checking Independence With Matrices

Turning the independence question into a homogeneous system and solving it with Week 2 tools.

8.1Turning the question into a system

To test $\{v_1, \dots, v_k\}$ we solve $\alpha_1 v_1 + \cdots + \alpha_k v_k = 0$ , a homogeneous system whose unknowns are the coefficients $\alpha_i$ . A homogeneous system is always consistent ( $\alpha = 0$ works), so the only question is whether the trivial solution is the only one.

only the trivial solution $\Rightarrow$ independent;
a non-zero solution exists (infinitely many) $\Rightarrow$ dependent.

Put the vectors as the columns of a matrix $V$ (column $j$ is $v_j$ ). The set is independent exactly when $V\mathbf{x} = \mathbf{0}$ has only the trivial solution.

Figure 11. The matrix test for independence

The determinant shortcut

If $V$ is square, just check $\det(V)$ . If $\det(V) \neq 0$ , $V$ is invertible and $V\mathbf{x} = \mathbf{0} \Rightarrow \mathbf{x} = V^{-1}\mathbf{0} = \mathbf{0}$ : independent. If $\det(V) = 0$ , Gauss elimination leaves a free variable: dependent.

8.2Too many vectors

Rule

Any set of $k$ vectors in $\mathbb{R}^n$ with $k \geq n + 1$ is linearly dependent. For instance, any 3 vectors in $\mathbb{R}^2$ , or any 4 vectors in $\mathbb{R}^3$ . The homogeneous system has $k$ unknowns but only $n$ equations, so Gauss elimination always leaves at least one free variable, and hence a non-zero solution.

8.3Going deeper

Count pivots. Independent $\iff$ every column of $V$ has a pivot. A $3 \times 4$ matrix has at most 3 pivots, so one of four columns is always free (extra phrasing of the rule above).
Fewer vectors than the dimension may still be dependent. $\{(1, 2), (3, 6)\}$ has two vectors in $\mathbb{R}^2$ and is dependent, because $\det = 0$ (extra).
Link with Week 2. This is exactly the homogeneous-system analysis of Gauss elimination: a free variable $\iff$ infinitely many solutions $\iff$ dependence.

8.4Exam tips and common mistakes

Exam tips

Count first: more vectors than components means dependent immediately.
Equal numbers: use the determinant. Fewer: use Gauss elimination.
Put the vectors as columns (rows also works for rank, but be consistent).

Common mistakes

Using the determinant on a non-square matrix.
Concluding "independent" because the augmented system is consistent; a homogeneous system is always consistent.

Practice questions (graded-assignment style)

Q1Four vectors in the plane

Multiple Choice Question (MCQ): choose the one correct option.

What can be said about the linear independence of a set of 4 vectors in $\mathbb{R}^2$ ?

It is always independent

It is always dependent

It is independent iff det = 0

Nothing can be said

Solution

Idea

Count unknowns and equations.

The system $\sum \alpha_j v_j = 0$ has 4 unknowns but only 2 equations.
At most 2 pivots, so at least $4 - 2 = 2$ free variables: a non-zero solution exists. Always dependent (rule: $k \geq n + 1$ , $4 \geq 3$ ).
Answer: option 2.

Q2Which v keeps {(2,3,0),(0,1,0),v} dependent?

Multiple Select Question (MSQ): select all that apply.

Suppose $\{(2, 3, 0), (0, 1, 0), v\}$ is a linearly dependent set in $\mathbb{R}^3$ . Which are possible candidates for $v$ ?

(2, 3, 1)

(1, 2, 0)

(1, 3, 0)

(0, 1, 1)

(\pi, e, 0)

Solution

Idea

Take $v = (p, q, r)$ and compute the determinant with the vectors as columns.

The third row is $(0, 0, r)$ ; expanding along it: $\det = r\cdot\det\begin{bmatrix}2&0\\3&1\end{bmatrix} = 2r$ .
Dependent $\iff 2r = 0 \iff r = 0$ : $v$ must lie in the $xy$ -plane spanned by the other two.
So options 2, 3, 5 (third entry 0) work; options 1 and 4 have $r = 1$ : independent.
E.g. $(1, 2, 0) = \tfrac12(2, 3, 0) + \tfrac12(0, 1, 0)$ .
Answer: options 2, 3 and 5.

Q3Count the values

Numerical Answer Type (NAT).

For how many values of $a$ is $\{(1, 2, a),\ (a, 0, a),\ (-1, a^2, 8),\ (0, 1, -3a)\}$ linearly independent in $\mathbb{R}^3$ ?

Solution

Idea

Count before computing.

There are 4 vectors in $\mathbb{R}^3$ , so $k = 4 \geq 3 + 1$ .
The system has 3 equations in 4 unknowns, so at least one is free whatever $a$ is.
So the set is dependent for every $a$ ; the number of values making it independent is 0.
Answer: 0

Q4A determinant condition

Numerical Answer Type (NAT).

For which $t$ is $\{(1, 2),\ (3, t)\}$ linearly dependent?

Solution

Idea

Set $\det\begin{bmatrix}1&3\\2&t\end{bmatrix} = 0$ .

$1\cdot t - 3\cdot 2 = 0$ , so $t = 6$ .
Then $(3, 6) = 3(1, 2)$ .
Answer: 6

Q5Three vectors in R³

Multiple Choice Question (MCQ): choose the one correct option.

The vectors $(1, 1, 1), (1, 2, -1), (0, 1, 0)$ form a matrix with determinant 2. The set is:

Linearly dependent

Linearly independent

Empty

Cannot be decided

Solution

Idea

Non-zero determinant means invertible.

$\det \neq 0$ , so $V\mathbf{x} = 0$ has only $\mathbf{x} = 0$ : independent.
Answer: option 2.

Revision overview

Set upVectors as columns of V
Solve V x = 0

OutcomeOnly x = 0: independent
Free variable: dependent

Square Vdet(V) ≠ 0: independent
det(V) = 0: dependent

Checking with matricesWeek 3 · Part 8

Too manyk ≥ n + 1 vectors in ℝⁿ
are always dependent

PivotsPivot in every column
⟺ independent

Always consistentHomogeneous system has
the trivial solution

Revision: independence as a question about Vx = 0

Next · Building blocksSpan and Spanning Sets

Building blocks

Span and Spanning Sets

Everything a set can build, and how to grow a set until it builds a whole space.

9.1The span of a set

Definition: span

The span of a set $S$ , written $\operatorname{span}(S)$ , is the set of all finite linear combinations of elements of $S$ :

\operatorname{span}(S) = \left\{ \sum_{i=1}^{n} \alpha_i v_i \;:\; v_i \in S,\ \alpha_i \in \mathbb{R} \right\}

Roughly, it is everything you can build from $S$ .

Example 37The span is always a subspace

Verify that $\operatorname{span}(S)$ is a subspace, for non-empty $S$ .

Solution

Non-empty. $\operatorname{span}(S)$ contains every $v \in S$ , since $v = 1\cdot v$ .
Addition. Let $x = \sum \alpha_i v_i$ and $y = \sum \beta_j w_j$ be in the span. Then $x + y$ is again a finite combination of elements of $S$ (list all the $v_i$ and $w_j$ together).
Scalars. $cx = \sum (c\alpha_i)v_i$ , again a finite combination.
Both closures hold, so by the subspace test $\operatorname{span}(S)$ is a subspace. $\blacksquare$

9.2Spanning sets

Definition: spanning set

Let $V$ be a vector space and $S \subseteq V$ . $S$ is a spanning set for $V$ (or $S$ spans $V$ ) if $\operatorname{span}(S) = V$ : every vector of $V$ can be built from $S$ .

Example 40Spanning sets of R²

Show that $\{(1, 0), (0, 1)\}$ , $\{(1, 0), (0, 1), (1, 1)\}$ and $\{(1, 1), (1, 0)\}$ all span $\mathbb{R}^2$ .

Solution

$\{(1, 0), (0, 1)\}$ : any $(x, y) = x(1, 0) + y(0, 1)$ .
$\{(1, 0), (0, 1), (1, 1)\}$ : $(x, y) = x(1, 0) + y(0, 1) + 0\cdot(1, 1)$ . Every superset of a spanning set is a spanning set.
$\{(1, 1), (1, 0)\}$ : $(x, y) = y(1, 1) + (x - y)(1, 0)$ . Check: $y(1, 1) + (x - y)(1, 0) = (y + x - y,\ y) = (x, y)$ . $\checkmark$

9.3Two facts about span

Facts

$T \subseteq S \Rightarrow \operatorname{span}(T) \subseteq \operatorname{span}(S)$ . A combination of vectors of $T$ is also a combination of vectors of $S$ . The inclusion can be strict: $\{(1, 0)\}$ spans the $x$ -axis, $\{(1, 0), (0, 1)\}$ spans $\mathbb{R}^2$ .
$\operatorname{span}(\operatorname{span}(S)) = \operatorname{span}(S)$ . Let $W = \operatorname{span}(S)$ . Each $w \in W$ equals $1\cdot w$ , so $W \subseteq \operatorname{span}(W)$ ; and combinations of vectors of $W$ stay in $W$ because $W$ is a subspace, so $\operatorname{span}(W) \subseteq W$ .

9.4Building a spanning set step by step

Grow a set one vector at a time, each time adding a vector that is not in the span of what you already have. For $\mathbb{R}^3$ :

Step	Set	Its span
1	$\emptyset$	$\{0\}$ (the empty combination is $0$ ), the smallest subspace
2	$\{(1, 1, 0)\}$	a line through the origin, not all of ℝ³
3	$\{(1, 1, 0), (1, 1, 1)\}$	the plane $\{(a + b, a + b, b)\}$ , $x = y$ , still not $\mathbb{R}^3$
4	$\{(1, 1, 0), (1, 1, 1), (1, 0, 0)\}$	all of $\mathbb{R}^3$

Verification of step 4: we need $a(1,1,0) + b(1,1,1) + c(1,0,0) = (a + b + c,\ a + b,\ b) = (x, y, z)$ . Reading from the last entry up: $b = z$ , $a = y - z$ , $c = x - y$ . A solution exists for every $(x, y, z)$ , so $S_3$ spans $\mathbb{R}^3$ . At each step the new vector was outside the previous span: $(1, 1, 1)$ is not on the line through $(1, 1, 0)$ , and $(1, 0, 0)$ has $x \neq y$ , so it is off the plane $x = y$ .

Figure 12. The span grows: point, line, plane, whole space

Example 41Construct a spanning set from {(1, 2, 3)}

Can you construct a spanning set for $\mathbb{R}^3$ starting with $\{(1, 2, 3)\}$ ?

Solution

Start. $S_1 = \{(1, 2, 3)\}$ spans the line $\{t(1, 2, 3)\}$ , not all of $\mathbb{R}^3$ .
Add a vector off that line. $(1, 0, 0)$ is not a multiple of $(1, 2, 3)$ . Let $S_2 = \{(1, 2, 3), (1, 0, 0)\}$ : its span is $\{(a + b,\ 2a,\ 3a)\}$ .
Identify that plane. Here $y = 2a$ and $z = 3a$ , so $3y = 6a = 2z$ : every point satisfies $3y - 2z = 0$ , a plane, still not $\mathbb{R}^3$ .
Add a vector off the plane. $(0, 1, 0)$ : $3(1) - 2(0) = 3 \neq 0$ , so it is not in $\operatorname{span}(S_2)$ . Let $S_3 = \{(1, 2, 3), (1, 0, 0), (0, 1, 0)\}$ .
Verify. Solve $a(1,2,3) + b(1,0,0) + c(0,1,0) = (a + b,\ 2a + c,\ 3a) = (x, y, z)$ : $a = \tfrac z3$ , $b = x - \tfrac z3$ , $c = y - \tfrac{2z}{3}$ . A solution exists for every $(x, y, z)$ , so $\operatorname{span}(S_3) = \mathbb{R}^3$ .
Spot-check with $(0, 0, 3)$ : $a = 1$ , $b = -1$ , $c = -2$ , and $1(1,2,3) - 1(1,0,0) - 2(0,1,0) = (0, 0, 3)$ . $\checkmark$

Answer: yes, for example $\{(1,2,3), (1,0,0), (0,1,0)\}$ .

9.5Going deeper

Span answers "which vectors can I reach?" A set spans $V$ iff the system " $\sum \alpha_i v_i = v$ " is solvable for every $v \in V$ .
Spanning and independence pull in opposite directions. Adding vectors keeps spanning but may destroy independence; removing vectors keeps independence but may destroy spanning. A set that does both well will be a basis (next chapter).
Removing a vector from a spanning set can break spanning. $\{(1, 0), (0, 1)\}$ spans $\mathbb{R}^2$ , but $\{(0, 1)\}$ spans only the $y$ -axis.

9.6Vector spaces beyond $\mathbb{R}^n$ , and subspaces that come from a matrix

Graded questions on this week do not stay inside $\mathbb{R}^n$ . They ask about polynomials, matrices and functions as vectors, about the solution set $\{\mathbf{x} : A\mathbf{x} = \mathbf{0}\}$ , and about sums of subspaces. This section collects the facts and the one technique that handles all of them.

A. Polynomial spaces

Let $P_n$ be the set of all polynomials of degree at most $n$ , including the zero polynomial, with the usual addition and scalar multiplication. It is a vector space, and $P_2 = \{a + bx + cx^2\}$ is spanned by $\{1, x, x^2\}$ .

"Degree exactly $n$ " is not a vector space. It has no zero vector, and it is not closed under addition: $x^2 + (-x^2 + x) = x$ has degree 1.
A polynomial is the zero vector only when all its coefficients are 0. So $a + bx + cx^2 = 0$ (for all $x$ ) forces $a = b = c = 0$ .
Subspaces of $P_2$ : $\{p : p(1) = 0\}$ passes (if $p(1) = 0$ and $q(1) = 0$ , then $(p + q)(1) = 0$ and $(cp)(1) = 0$ ). $\{p : p(0) = 1\}$ fails (the zero polynomial has $p(0) = 0$ ). The even polynomials $p(x) = p(-x)$ pass and equal $\operatorname{span}\{1, x^2\}$ .

B. Matrix spaces and function spaces

$M_{m \times n}$ with entrywise operations is a vector space; the zero vector is the zero matrix. Subspaces of $M_{2 \times 2}$ : symmetric matrices, upper triangular matrices, matrices with trace $0$ . Not subspaces: invertible matrices, $\{A : \det A = 0\}$ ( $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = I$ has $\det = 1$ ), matrices with integer entries (not closed under scalars like $\tfrac12$ ).
Real functions $f : \mathbb{R} \to \mathbb{R}$ , with $(f + g)(x) = f(x) + g(x)$ and $(cf)(x) = c\,f(x)$ , form a vector space. The zero vector is the function that is $0$ for every $x$ . $\{f : f(0) = 0\}$ is a subspace; $\{f : f(1) = 1\}$ is not.
Dependence of functions. $f_1, \dots, f_k$ are dependent if some non-trivial combination is zero for every $x$ . To prove independence, plug in a few values of $x$ . $\{\sin x, \cos x\}$ : $c_1\sin x + c_2\cos x = 0$ at $x = 0$ gives $c_2 = 0$ , and at $x = \pi/2$ gives $c_1 = 0$ , so independent. To prove dependence, exhibit the relation: $\{1, \cos 2x, \sin^2 x\}$ is dependent because $1 - \cos 2x - 2\sin^2 x = 0$ for every $x$ .

C. The one technique: turn it into coordinates

To test independence or spanning in $P_2$ or $M_{2 \times 2}$ , write each vector as a column of coefficients or entries, then use the matrix tests of this week:

Space	Vector	Coordinates
$P_2$	$a + bx + cx^2$	$(a, b, c) \in \mathbb{R}^3$
$M_{2 \times 2}$	$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$	$(a, b, c, d) \in \mathbb{R}^4$

Example. Do $1 + x,\; 1 - x,\; x^2$ span $P_2$ ? Their coordinates are $(1, 1, 0)$ , $(1, -1, 0)$ , $(0, 0, 1)$ . The determinant of the matrix with these rows is $1\cdot(-1) - 1\cdot 1 = -2 \neq 0$ , so the three are independent, and three independent vectors in a 3-coordinate space span it. Yes. By contrast $1 + x,\; 2 + 2x,\; x^2$ gives determinant $0$ (the second is twice the first), so it does not span.

Why this is legitimate. Adding polynomials adds coefficients, and scaling a polynomial scales coefficients, exactly as for vectors in

\mathbb{R}^3

. So every statement about combinations, independence and span means the same in both descriptions.

D. Subspaces that come from a matrix

Let $A$ be $m \times n$ .

Null space $N(A) = \{\mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \mathbf{0}\}$ , the solution set of the homogeneous system. It is a subspace: $A\mathbf{0} = \mathbf{0}$ ; and if $A\mathbf{x} = \mathbf{0}$ , $A\mathbf{y} = \mathbf{0}$ , then $A(\mathbf{x} + \mathbf{y}) = \mathbf{0}$ and $A(c\mathbf{x}) = \mathbf{0}$ .
The solution set of $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \neq \mathbf{0}$ is not a subspace: it does not contain $\mathbf{0}$ (this is the "shifted plane" of Part 3).
Column space $C(A) = \operatorname{span}\{\text{columns of } A\} \subseteq \mathbb{R}^m$ . Since $A\mathbf{x} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n$ , a vector $\mathbf{b}$ is in $C(A)$ exactly when $A\mathbf{x} = \mathbf{b}$ has a solution. A spanning set is the pivot columns of $A$ (the columns of the original $A$ that hold the leading 1s of its RREF).
Row space is the span of the rows of $A$ , a subspace of $\mathbb{R}^n$ . Row operations do not change it.

Worked example. $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 3 \\ 3 & 6 & 4 \end{bmatrix}$ has RREF $\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ with pivots in columns 1 and 3.

Null space: $x_1 + 2x_2 = 0$ , $x_3 = 0$ , and $x_2 = t$ is free. So $\mathbf{x} = t(-2, 1, 0)$ and $N(A) = \operatorname{span}\{(-2, 1, 0)\}$ . Check: $A(-2, 1, 0)^T = (-2 + 2,\; -4 + 4,\; -6 + 6) = \mathbf{0}$ ✓.
Column space: $C(A) = \operatorname{span}\{(1, 2, 3),\; (1, 3, 4)\}$ , the original columns 1 and 3. Column 2, $(2, 4, 6)$ , is twice column 1, so it adds nothing.
Is $(2, 5, 7)$ in $C(A)$ ? It equals $(1, 2, 3) + (1, 3, 4)$ : yes. Is $(1, 0, 0)$ ? We would need $c_1 + c_2 = 1$ and $2c_1 + 3c_2 = 0$ , giving $c_2 = -2$ , $c_1 = 3$ ; the third entry gives $3\cdot 3 + (-2)\cdot 4 = 1 \neq 0$ . No.

E. Sum and intersection of subspaces

$U \cap W$ is always a subspace (see the subspace chapter).
$U \cup W$ is a subspace only when one of them contains the other.
The sum $U + W = \{u + w : u \in U,\ w \in W\}$ is always a subspace, and it is the smallest subspace containing both. Also $\operatorname{span}(S) + \operatorname{span}(T) = \operatorname{span}(S \cup T)$ .
Example in $\mathbb{R}^2$ : $U$ = the $x$ -axis, $W$ = the $y$ -axis. $U \cap W = \{\mathbf{0}\}$ , $U \cup W$ is not a subspace (it holds $(1, 0)$ and $(0, 1)$ but not $(1, 1)$ ), and $U + W = \mathbb{R}^2$ .

9.7Exam tips and common mistakes

Exam tips

To show a set spans, take a general vector (x, y, z) and solve for the coefficients.
To show it does not span, find one vector you cannot reach (e.g. an equation all combinations satisfy, and a vector that breaks it).

Common mistakes

Believing that a spanning set cannot lose a vector. It can fail to span after removal.
Confusing span (a set of vectors) with the spanning set (the generators).

Practice questions (graded-assignment style)

Q1Adding and removing

Multiple Select Question (MSQ): select all that apply.

Choose the correct options.

If S spans V, then S ∪ {v} spans V for all v ∈ V.

The span of the empty set is the zero vector space.

If S spans V, then S \ {v} spans V for all v ∈ S.

If S spans V, then S ∪ {v} may fail to span V for some v ∈ V.

Solution

Idea

Use span(S) ⊆ span(S ∪ {v}) ⊆ V.

Option 1: $V = \operatorname{span}(S) \subseteq \operatorname{span}(S \cup \{v\}) \subseteq V$ , so equality holds.
Option 2: the empty combination is $0$ , so $\operatorname{span}(\emptyset) = \{0\}$ .
Option 3 fails: $\{(1,0),(0,1)\}$ spans $\mathbb{R}^2$ , but removing $(1,0)$ leaves $\operatorname{span}\{(0,1)\} = \{(0, t)\}$ , the $y$ -axis. Option 4 contradicts Option 1.
Answer: options 1 and 2.

Q2The span of three matrices

Multiple Select Question (MSQ): select all that apply.

Let $S = \left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$ . Choose the correct statement(s).

span(S) is the space of lower triangular 2 × 2 matrices

span(S) is the space of upper triangular 2 × 2 matrices

span(S) is the space of all 2 × 2 matrices

span(S) is the space of scalar 2 × 2 matrices

Solution

Idea

Write a general combination a·E₁₁ + b·E₁₂ + c·E₂₂.

$a E_{11} + b E_{12} + c E_{22} = \begin{bmatrix}a&b\\0&c\end{bmatrix}$ : the entry below the diagonal is always 0, and $a, b, c$ are arbitrary. That is the upper triangular matrices (option 2).
Option 1: $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is in the span but not lower triangular. Option 3: a matrix with a non-zero entry below the diagonal is never reached. Option 4: scalar matrices $kI$ are only a small part; $\begin{bmatrix}1&5\\0&2\end{bmatrix}$ is in the span but not scalar.
Answer: option 2.

Q3A coefficient

Numerical Answer Type (NAT).

Write $(4, 5, 3) = a(1, 2, 3) + b(1, 0, 0) + c(0, 1, 0)$ . Find $b$ .

Solution

Idea

Use $a = z/3$ , $b = x - z/3$ , $c = y - 2z/3$ .

$3a = 3$ , so $a = 1$ .
$a + b = 4$ , so $b = 3$ ; $2a + c = 5$ , so $c = 3$ .
Check: $1(1,2,3) + 3(1,0,0) + 3(0,1,0) = (4, 5, 3)$ .
Answer: 3

Q4Does it span?

Multiple Choice Question (MCQ): choose the one correct option.

Does $\{(1, 2), (2, 4)\}$ span $\mathbb{R}^2$ ?

Yes, any two vectors span ℝ²

No: every combination is a multiple of

(1, 2)

Yes, because it contains (2, 4)

No, because it has two vectors

Solution

Idea

Is one a multiple of the other?

$(2, 4) = 2(1, 2)$ , so $a(1,2) + b(2,4) = (a + 2b)(1, 2)$ : only the line $y = 2x$ . Not all of $\mathbb{R}^2$ ; e.g. $(1, 0)$ is not reached.
Answer: option 2.

Q5Properties of span

Multiple Select Question (MSQ): select all that apply.

Which statements are true for subsets $T \subseteq S$ of a vector space?

\operatorname{span}(T) \subseteq \operatorname{span}(S)

\operatorname{span}(\operatorname{span}(S)) = \operatorname{span}(S)

\operatorname{span}(S)

is a subspace

\operatorname{span}(T) = \operatorname{span}(S)

always

Solution

Idea

Recall the two facts and the verification.

Options 1, 2, 3 are proved in this part.
Option 4 is false: $T = \{(1,0)\}\subseteq S = \{(1,0),(0,1)\}$ have different spans.
Answer: options 1, 2 and 3.

Q1Subspaces of the polynomials of degree at most 2

Multiple Select Question (MSQ): select all that apply.

Which of these are subspaces of $P_2$ , the space of polynomials of degree at most 2?

\{p : p(1) = 0\}

\{p : p(0) = 1\}

polynomials of degree exactly 2

\{p : p(x) = p(-x)\ \text{for all } x\}

\{p : p(0) = p(1)\}

Solution

Idea

Write $p = a + bx + cx^2$ . A set cut out by homogeneous linear conditions on $a, b, c$ is a subspace; anything else is hunted for a failed test.

1 ✓ $p(1) = a + b + c = 0$ is homogeneous linear.
2 ✗ $p(0) = a = 1$ is not homogeneous; the zero polynomial has $p(0) = 0$ , so $\mathbf{0}$ is missing.
3 ✗ No zero polynomial (degree of $0$ is not 2), and not closed: $x^2 + (-x^2 + x) = x$ .
4 ✓ $p(-x) = a - bx + cx^2$ , so the condition is $b = 0$ : homogeneous linear.
5 ✓ $p(0) = a$ and $p(1) = a + b + c$ , so the condition is $b + c = 0$ : homogeneous linear.
Answer: options 1, 4 and 5.

Q2Dependent polynomials

Numerical Answer Type (NAT).

For which value of $k$ are the polynomials $1 + x + x^2$ , $2 + 3x + 4x^2$ and $1 + 2x + kx^2$ linearly dependent?

Solution

Idea

Replace each polynomial by its coefficient vector; three vectors in $\mathbb{R}^3$ are dependent exactly when the determinant is $0$ .

Coefficient vectors: $(1, 1, 1)$ , $(2, 3, 4)$ , $(1, 2, k)$ .
$\det\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & k \end{bmatrix} = 1(3k - 8) - 1(2k - 4) + 1(4 - 3) = k - 3$ .
$k - 3 = 0 \Rightarrow k = 3$ . Check: $(2, 3, 4) - (1, 1, 1) = (1, 2, 3)$ , so the third polynomial $1 + 2x + 3x^2$ is the difference of the first two ✓.
Answer: 3

Q3Which set spans $P_2$ ?

Multiple Choice Question (MCQ): choose the one correct option.

Which of these sets spans $P_2$ ?

\{1, x\}

\{1 + x,\ 1 - x,\ x^2\}

\{1, x, 1 + x\}

\{x, x^2, x + x^2\}

Solution

Idea

$P_2$ needs three independent "directions" (coordinates in $\mathbb{R}^3$ ). Fewer than three vectors cannot span, and three vectors span only if they are independent.

1 ✗ Only two vectors, and nothing produces $x^2$ .
2 ✓ Coordinates $(1, 1, 0)$ , $(1, -1, 0)$ , $(0, 0, 1)$ have determinant $-2 \neq 0$ .
3 ✗ The third is the sum of the first two, so there are only two independent vectors, and no $x^2$ is ever produced.
4 ✗ The third is the sum of the first two, and the constant $1$ is never produced.
Answer: option 2.

Q4Subspaces of 2 × 2 matrices

Multiple Select Question (MSQ): select all that apply.

Which of these are subspaces of $M_{2 \times 2}$ ?

symmetric matrices

matrices with trace

0

matrices with

\det = 0

upper triangular matrices

matrices with all entries integers

Solution

Idea

Check the zero matrix, then closure under addition and under every real scalar.

1 ✓ $A^T = A$ and $B^T = B$ give $(A + B)^T = A + B$ and $(cA)^T = cA$ .
2 ✓ $\operatorname{tr}(A + B) = \operatorname{tr}A + \operatorname{tr}B = 0$ and $\operatorname{tr}(cA) = c\operatorname{tr}A = 0$ .
3 ✗ $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ have $\det = 0$ , but their sum $I$ has $\det = 1$ .
4 ✓ Sums and scalar multiples keep the lower-left entry $0$ .
5 ✗ $\tfrac12 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ has non-integer entries: not closed under real scalars.
Answer: options 1, 2 and 4.

Q5A vector in the null space

Multiple Choice Question (MCQ): choose the one correct option.

For $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 3 \\ 3 & 6 & 4 \end{bmatrix}$ , which vector satisfies $A\mathbf{x} = \mathbf{0}$ ?

(-2, 1, 0)

(2, 1, 0)

(1, -2, 0)

(0, 0, 1)

Solution

Idea

Solve $A\mathbf{x} = \mathbf{0}$ from the RREF, or just multiply each candidate.

RREF is $\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ : $x_1 = -2x_2$ , $x_3 = 0$ , $x_2$ free. All solutions are $t(-2, 1, 0)$ .
Option 1: $A(-2, 1, 0)^T = (0, 0, 0)$ ✓. Option 2 gives $(4, 8, 12)$ , option 3 gives $(-3, -6, -9)$ , option 4 gives the third column $(1, 3, 4)$ : none is $\mathbf{0}$ .
Answer: option 1.

Q6Which vectors lie in the column space?

Multiple Select Question (MSQ): select all that apply.

For the same $A$ , which vectors lie in the column space of $A$ ?

(2, 5, 7)

(1, 0, 0)

(0, 0, 0)

(1, 2, 3)

Solution

Idea

$\mathbf{b}$ is in the column space exactly when $A\mathbf{x} = \mathbf{b}$ is consistent. The column space is spanned by the pivot columns $(1, 2, 3)$ and $(1, 3, 4)$ .

1 ✓ $(2, 5, 7) = (1, 2, 3) + (1, 3, 4)$ .
2 ✗ $c_1 + c_2 = 1$ , $2c_1 + 3c_2 = 0$ give $c_1 = 3, c_2 = -2$ , but then the third entry is $9 - 8 = 1 \neq 0$ . (Equivalently, the rank of $[A \mid \mathbf{b}]$ is 3, bigger than $\operatorname{rank}A = 2$ .)
3 ✓ The zero vector is in every subspace.
4 ✓ It is the first column itself.
Answer: options 1, 3 and 4.

Q7Two axes in the plane

Multiple Select Question (MSQ): select all that apply.

Let $U$ be the $x$ -axis and $W$ the $y$ -axis in $\mathbb{R}^2$ . Choose the correct statements.

U \cap W = \{\mathbf{0}\}

U \cup W

is a subspace of

\mathbb{R}^2

U + W = \mathbb{R}^2

U + W = U \cup W

Solution

Idea

$U + W$ is every vector $u + w$ ; the union only has vectors lying on an axis.

1 ✓ Only the origin is on both axes.
2 ✗ $(1, 0) + (0, 1) = (1, 1)$ is on neither axis.
3 ✓ $(x, y) = (x, 0) + (0, y)$ .
4 ✗ $U + W = \mathbb{R}^2$ contains $(1, 1)$ ; $U \cup W$ does not.
Answer: options 1 and 3.

Q8A relation between functions

Numerical Answer Type (NAT).

The functions $\sin^2 x$ , $\cos^2 x$ and $1$ are linearly dependent: $a\sin^2 x + b\cos^2 x + c \cdot 1 = 0$ for every $x$ , with $a = b = 1$ . Find $c$ .

Solution

Idea

Use the identity $\sin^2 x + \cos^2 x = 1$ to find the combination that is zero for all $x$ .

$\sin^2 x + \cos^2 x - 1 = 0$ for every $x$ , so with $a = b = 1$ we need $c = -1$ .
Check at $x = 0$ : $0 + 1 + c = 0 \Rightarrow c = -1$ ✓. A non-trivial combination (not all coefficients zero) equal to the zero function means the set is dependent.
Answer: -1

Revision overview

SpanAll finite linear
combinations of S

Subspacespan(S) is always
a subspace

Spanning setspan(S) = V
Supersets also span

Span and spanning setsWeek 3 · Part 9

FactsT ⊆ S ⟹ span T ⊆ span S
span(span S) = span S

Build up{0}, line, plane, ℝ³
add a vector off the span

NextBasis: spanning and
independent together

Revision: what a set can build and how to grow it

Next · All partsWeek 3 Summary Sheet

All parts

Week 3 Summary Sheet

One page to revise all of Week 3: definitions, tests, and the mistakes to avoid.

The big picture

Part 1 · Why vector spaces

Homogeneous solution sets are closed under + and scalars.
Linear combinations of solutions are solutions.

Part 2 · Axioms

Nine axioms; one failure kills a vector space.
The zero vector need not be (0, …, 0).

Part 3 · Examples

$\mathbb{R}^n$ , matrices, solutions of $Ax = 0$ are vector spaces.
Strange operations: find the zero first.

Part 4 · Properties

Cancellation law; unique zero and inverses.
$0\cdot u = 0_V$ , $(-1)u = -u$ .

Part 5 · Subspaces

Non-empty, closed under + and scalars.
Lines and planes through $0$ ; intersections.

Part 6 · Combinations

$\sum \alpha_i v_i$ ; test by solving a system.
Parallelogram law in ℝ².

Part 7 · Dependence

Dependent: non-trivial combination gives 0.
Independent: only the trivial one.

Part 8 · Matrix test

Columns of $V$ ; solve $V\mathbf{x} = 0$ .
More than $n$ vectors in $\mathbb{R}^n$ : dependent.

Part 9 · Span

All combinations; always a subspace.
Spanning set: span = V.

Revision sheet

Topic	Key idea
Vector space	non-empty set with $+$ and scalar $\cdot$ satisfying all 9 axioms
Not a vector space	one counterexample to one axiom is enough
Zero vector	whatever satisfies $u + 0_V = u$ ; unique; equals $0\cdot u$ , not always $(0, \dots, 0)$
Additive inverse	unique; written $-u$ ; equals $(-1)\cdot u$
Cancellation law	$v_1 + v_3 = v_2 + v_3 \Rightarrow v_1 = v_2$
Scalar facts	$0\cdot u = 0_V$ , $c\cdot 0_V = 0_V$ , $(-c)u = -(cu) = c(-u)$
Subspace test	non-empty + closed under addition + closed under scalar multiplication
Quick subspace check	$0_V \notin W \Rightarrow$ not a subspace (the converse fails)
Lines and planes	a subspace iff it passes through the origin
Solutions of $Ax = b$	a subspace iff $b = 0$
Linearly dependent	$\sum \alpha_i v_i = 0$ with some $\alpha_i \neq 0$ ; any set containing $0$ is dependent
Linearly independent	$\sum \alpha_i v_i = 0 \Rightarrow$ all $\alpha_i = 0$
Two vectors	independent iff not scalar multiples
Matrix test	columns of $V$ ; independent iff $V\mathbf{x} = 0$ only trivially; square: $\det V \neq 0$
Too many vectors	$k \geq n + 1$ vectors in $\mathbb{R}^n$ are always dependent
Supersets / subsets	superset of dependent is dependent; subset of independent is independent
Span	all finite linear combinations; always a subspace; $\operatorname{span}(\emptyset) = \{0\}$
Spanning set	$\operatorname{span}(S) = V$ ; every superset of a spanning set spans
Span facts	$T \subseteq S \Rightarrow \operatorname{span}T \subseteq \operatorname{span}S$ ; $\operatorname{span}(\operatorname{span}S) = \operatorname{span}S$

Which test when?

Question	Method
Is (V, +, ·) a vector space?	find 0_V first; then test unity and distributivity; or hunt for a counterexample
Is W a subspace of V?	zero test, then closure under $+$ and scalars (try $-1$ for inequalities)
Is v a combination of v₁, …, vₙ?	solve Σαᵢvᵢ = v; inconsistent means no
Dependent or independent?	count first ( $k \geq n + 1$ ); square: $\det$ ; else Gauss elimination on $V\mathbf{x} = 0$
Does S span V?	take a general vector and solve for the coefficients

Common mistakes

Zero vector assumed

It is not always $(0, \dots, 0)$ . Find it from the operations.

Few axioms checked

All nine must hold; one failure is enough to reject.

0 ∈ W is enough

It is necessary only. The parabola has 0 and is not a subspace.

Zero combination proves independence

The all-zero combination always gives 0. Show it is the only one.

Determinant on non-square V

Use Gauss elimination or count pivots instead.

Removing keeps spanning

Removing a vector can break spanning; adding one never does.

Mathematics forData Science 2

Chapters

Why Vector Spaces? A Motivating System

1.1A motivating system

1.2The key observation: closure

1.3Seeing S concretely

1.4What is a vector space, roughly?

1.5A contrast: the non-homogeneous case

1.6Going deeper

1.7Exam tips and common mistakes

Practice questions (graded-assignment style)

The Vector Space Axioms

2.1The definition

2.2Things to keep in mind

2.3Exam tips and common mistakes

Practice questions (graded-assignment style)

Is It a Vector Space? Worked Examples

3.1The standard examples

3.2Strange operations

3.3Going deeper

3.4Exam tips and common mistakes

Practice questions (graded-assignment style)

Properties of Vector Spaces

4.1The cancellation law

4.2Uniqueness of zero and inverses

4.3Scalars meeting zero

4.4Going deeper

4.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Subspaces

5.1The definition and the two-step test

5.2Examples that fail

5.3Lines, systems and half-spaces: three proofs

5.4Going deeper

5.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Linear Combinations

6.1Definition

6.2A picture in the plane

6.3A picture in space

6.4How to test "is v a combination?"

6.5Going deeper

6.6Exam tips and common mistakes

Practice questions (graded-assignment style)

Linear Dependence and Independence

7.1Linear dependence

7.2Linear independence

7.3Two vectors, three vectors

7.4Going deeper

7.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Checking Independence With Matrices

8.1Turning the question into a system

8.2Too many vectors

8.3Going deeper

8.4Exam tips and common mistakes

Practice questions (graded-assignment style)

Span and Spanning Sets

9.1The span of a set

9.2Spanning sets

9.3Two facts about span

9.4Building a spanning set step by step

9.5Going deeper

9.6Vector spaces beyond Rn\mathbb{R}^nRn, and subspaces that come from a matrix

A. Polynomial spaces

B. Matrix spaces and function spaces

C. The one technique: turn it into coordinates

D. Subspaces that come from a matrix

E. Sum and intersection of subspaces

9.7Exam tips and common mistakes

Practice questions (graded-assignment style)

Week 3 Summary Sheet

The big picture

Revision sheet

Which test when?

Common mistakes

Mathematics for
Data Science 2

9.6Vector spaces beyond $\mathbb{R}^n$ , and subspaces that come from a matrix