Linear-Algebra-Done-Right · Unit 2 · Video 1 · Interactive Practice

Definition of a Vector Space: Fⁿ, F^∞ and Function Spaces

IKey Formulas

Formula	Name	What it says
$a(u + v) = au + av$ , $\ (a + b)v = av + bv$	Distributive properties	In $(a + b)v$ the left $+$ adds scalars in $\mathbf{F}$ ; the right $+$ is the addition on $V$
$(x_1, x_2, \dots) + (y_1, y_2, \dots) = (x_1 + y_1, x_2 + y_2, \dots)$ , $\ \lambda(x_1, x_2, \dots) = (\lambda x_1, \lambda x_2, \dots)$	Operations on $\mathbf{F}^\infty$	Coordinate by coordinate; the additive identity is $(0, 0, 0, \dots)$
$(f + g)(x) = f(x) + g(x)$ , $\ (\lambda f)(x) = \lambda f(x)$	Operations on $\mathbf{F}^S$	Point by point, for all $x \in S$
$0(x) = 0$ , $\ (-f)(x) = -f(x)$	Identity and inverse in $\mathbf{F}^S$	The zero function and the negative of $f$ , for all $x \in S$

Key Insight: A vector space is a set $V$ with an addition and a scalar multiplication on $V$ satisfying six properties (commutativity, associativity, additive identity, additive inverse, multiplicative identity, distributive properties) and nothing more, so its vectors may be lists, sequences or functions.

IIVisualization 1 — Testing the Six Properties

A set with two operations is a vector space only if all six properties hold for every choice of vectors and scalars.

Operations on R² a 2 b −1

💡 A check at chosen values can refute a property but never prove it; a proof must cover all $u, v, w \in V$ and all $a, b \in \mathbf{F}$ .

IIIVisualization 2 — Pointwise Operations in $\mathbf{R}^{[0,1]}$

In $\mathbf{R}^{[0,1]}$ the vectors are functions, and sums and scalar multiples are computed one point $x$ at a time.

Operation x 0.5 λ 2

IVVisualization 3 — Lists as Functions

A list $x \in \mathbf{F}^n$ is the function $k \mapsto x_k$ on $\{1, \dots, n\}$ , so coordinate addition is pointwise addition.

Space k 2

VQuiz Questions

Problem 1 · A Coordinate in R^∞

Given: $x = (1, 2, 3, \dots)$ with $x_k = k$ , and $y = (1, -1, 1, -1, \dots)$ with $y_k = (-1)^{k+1}$ , both in $\mathbf{R}^\infty$ — find the fourth coordinate of $2x + 3y$ .

✅ Correct! Operations on

\mathbf{R}^\infty

act coordinate by coordinate:

2x_4 + 3y_4 = 2 \cdot 4 + 3 \cdot (-1) = 5

❌ Check the sign of $y_4$ .

y_4 = (-1)^{4+1} = (-1)^5 = -1

, not

1

❌ That is the third coordinate.

2x_3 + 3y_3 = 6 + 3 = 9

; the fourth uses

x_4 = 4

and

y_4 = -1

❌ The scalars do not combine like that.

15 = (2 + 3)(4 - 1)

, but the coordinate is

2x_4 + 3y_4

: each scalar multiplies its own sequence.

❌ Not quite. Compute

2x_4

and

3y_4

separately, then add.

Show solution

Scalar multiplication and addition in $\mathbf{R}^\infty$ work coordinate by coordinate, so the $k^\text{th}$ coordinate of $2x + 3y$ is $2x_k + 3y_k$ .

x_4 = 4, \qquad y_4 = (-1)^{5} = -1

2x_4 + 3y_4 = 2 \cdot 4 + 3 \cdot (-1) = 8 - 3 = 5

Indeed $2x + 3y = (5, 1, 9, 5, 13, \dots)$ , whose fourth coordinate is $\mathbf{5}$ .

Problem 2 · A Hidden Failure

Given: $V = \mathbf{R}^2$ with the usual addition but the scalar multiplication $\lambda(x_1, x_2) = (\lambda x_1, x_2)$ — which property fails?

✅ Correct!

(a + b)v = ((a + b)x_1, x_2)

but

av + bv = ((a + b)x_1, 2x_2)

: the second coordinate is counted twice.

❌ That one holds.

1(x_1, x_2) = (1 \cdot x_1, x_2) = (x_1, x_2)

❌ That one holds. Both sides equal

(abx_1, x_2)

, since the second coordinate is never touched.

❌ That one holds.

a(u + v) = (a(u_1 + v_1), u_2 + v_2)

and

au + av = (au_1 + av_1, u_2 + v_2)

agree.

❌ Not quite. Compute both sides of each property in the second coordinate.

Show solution

Addition is the usual one, so commutativity, associativity of addition, the additive identity $(0, 0)$ and additive inverses all hold. Check the properties that involve scalars, with $u = (u_1, u_2)$ and $v = (x_1, x_2)$ :

$1v = (x_1, x_2) = v$ ✓
$(ab)v = (abx_1, x_2)$ and $a(bv) = a(bx_1, x_2) = (abx_1, x_2)$ ✓
$a(u + v) = (a(u_1 + x_1), u_2 + x_2) = (au_1, u_2) + (ax_1, x_2) = au + av$ ✓
$(a + b)v = ((a + b)x_1, x_2)$ , but $av + bv = (ax_1, x_2) + (bx_1, x_2) = ((a + b)x_1, 2x_2)$ ✗

Counterexample: $a = b = 1$ , $v = (0, 1)$ gives $2v = (0, 1)$ but $v + v = (0, 2)$ . So $V$ is not a vector space, and only the second distributive property fails.

Problem 3 · Functions on a Three-Point Set

Given: $S = \{1, 2, 3\}$ and $f, g \in \mathbf{R}^S$ with $f(k) = k^2 - 2$ and $g(k) = 3 - k$ — find the values of $h = 2f + (-1)g$ and the additive inverse of $f$ .

What is $(h(1), h(2), h(3))$ ?

Which function is $-f$ ?

✅ Correct!

h = (-4, 3, 14)

point by point, and

f(k) + (2 - k^2) = 0

at every

k \in S

, so

-f

k \mapsto 2 - k^2

❌ That is $f - g$ . The scalar

2

multiplies every value of

f

2f = (-2, 4, 14)

❌ That is $2f + g$ . The scalar

-1

turns

g = (2, 1, 0)

into

(-2, -1, 0)

❌ That is $2(f - g)$ . Only

f

is doubled;

g

is multiplied by

-1

❌ That is a multiplicative inverse. The additive inverse must give

f(k) + h(k) = 0

; at

k = 1

-1 + (-1) = -2 \ne 0

❌ Check the constant term.

(k^2 - 2) + (-k^2 - 2) = -4

, not

0

: negate both terms.

❌ That is the additive identity.

f + 0 = f

, and

f

is not the zero function, since

f(1) = -1

; the inverse must give

f + (-f) = 0

❌ Check $h$ . Evaluate

2f(k) - g(k)

k = 1, 2, 3

❌ Check $-f$ . It is defined by

(-f)(k) = -f(k)

Show solution

Tabulate: $f = (f(1), f(2), f(3)) = (-1, 2, 7)$ and $g = (2, 1, 0)$ . The operations are pointwise, $h(k) = 2f(k) + (-1)g(k)$ :

h(1) = 2(-1) - 2 = -4, \quad h(2) = 2(2) - 1 = 3, \quad h(3) = 2(7) - 0 = 14

In formula form $h(k) = 2(k^2 - 2) - (3 - k) = 2k^2 + k - 7$ , which gives the same three values.

The additive inverse is defined by $(-f)(k) = -f(k) = -(k^2 - 2) = 2 - k^2$ . Check: $(f + (-f))(k) = (k^2 - 2) + (2 - k^2) = 0 = 0(k)$ for every $k \in S$ , so $f + (-f)$ is the zero function.

Problem 4 · A Vector Space of Positive Numbers

Given: $V = (0, \infty)$ with addition $u \oplus v = uv$ and scalar multiplication $\lambda \odot v = v^\lambda$ for $\lambda \in \mathbf{R}$ ; with these operations $V$ is a real vector space — find its additive identity and the additive inverse of $5$ .

What is the additive identity of $V$ ?

What is the additive inverse of $5$ ?

✅ Correct!

v \oplus 1 = v \cdot 1 = v

, so the zero vector of

V

is the number

1

, and

5 \oplus \tfrac{1}{5} = 1

❌ $0$ is not in $V = (0, \infty)$ . The zero vector is whatever element

z \in V

satisfies

v \oplus z = vz = v

❌ The zero vector need not be the number $0$ . Solve

v \oplus z = v

, that is

vz = v

, inside

V

❌ Test it.

v \oplus e = ev \ne v

; for instance

2 \oplus e = 2e

. The zero vector

z

must satisfy

v \oplus z = vz = v

❌ $0$ is not in $V = (0, \infty)$ . The inverse

w

must satisfy

5 \oplus w = 5w = 1

, the zero vector of

V

❌ $-5$ is not in $V$ . Also

5 \oplus (-5) = -25

, which is not the additive identity.

❌ That is the additive identity.

5 \oplus 1 = 5

; the inverse

w

must give

5 \oplus w = 1

❌ Check the identity. Solve

vz = v

for

z

❌ Check the inverse. Solve

5 \oplus w = 5w = 1

Show solution

Identity. We need $z \in V$ with $v \oplus z = v$ for all $v \in V$ , that is $vz = v$ . Dividing by $v > 0$ gives $z = 1 \in V$ .

Inverse. We need $w \in V$ with $5 \oplus w = 1$ (the zero vector of $V$ ), that is $5w = 1$ , so $w = \tfrac{1}{5} \in V$ . In general the additive inverse of $v$ is $\tfrac{1}{v}$ .

The other properties are laws of exponents: $uv = vu$ , $(uv)w = u(vw)$ , $\big(v^b\big)^a = v^{ab}$ , $v^1 = v$ , $(uv)^a = u^a v^a$ and $v^{a + b} = v^a v^b$ . For instance $2 \odot (3 \oplus 4) = 12^2 = 144 = 9 \cdot 16 = (2 \odot 3) \oplus (2 \odot 4)$ .

Solved: 0 / 4

Definition of a Vector Space: Fⁿ, F^∞ and Function Spaces

IKey Formulas

IIVisualization 1 — Testing the Six Properties

IIIVisualization 2 — Pointwise Operations in R[0,1]\mathbf{R}^{[0,1]}R[0,1]

IVVisualization 3 — Lists as Functions

VQuiz Questions

IIIVisualization 2 — Pointwise Operations in $\mathbf{R}^{[0,1]}$