Vector Spaces and Isomorphism

Move the coefficients and watch a line and a point move together. Spaces with different looks share one structure.

A vector space supports adding and scaling. Pairs of numbers, polynomials, and functions all obey the same rules. Same rules means essentially the same world. Calculations transfer from one side to the other.

For example, write a recipe on a card. The card's numbers rebuild the dish exactly. A line, too, is captured by two numbers: slope and intercept. Card and line move as one.

Examples & Exercises

Show that the set $S$ of all real sequences, with termwise addition $(x+y)_n = x_n + y_n$ and scalar multiplication $(cx)_n = cx_n$ , is a vector space.

(x+y)_n = x_n + y_n,\qquad (cx)_n = cx_n

Sliders

a = 1

b = 0.5

— the line

y = 1 + 0.5x

and the point

(1, 0.5)

Show proof

Closure: if $x, y \in S$ then every term of $x + y = (x_n + y_n)$ is real, so $x + y \in S$ . Likewise every term of $cx = (cx_n)$ is real, so $cx \in S$ .
$(S, +)$ is an abelian group: commutativity and associativity are inherited termwise from $\mathbb{R}$ , the identity is the zero sequence $(0, 0, \dots)$ , and the inverse of $x$ is $(-x_n)$ .
The remaining scalar axioms hold termwise too: $(c(dx))_n = c(dx_n) = (cd)x_n = ((cd)x)_n$ , and $(1x)_n = 1 \cdot x_n = x_n$ .
The distributive laws check out at the sequence level: $(c(x+y))_n = c(x_n + y_n) = cx_n + cy_n = (cx + cy)_n$ , and $((c+d)x)_n = (c+d)x_n = cx_n + dx_n = (cx + dx)_n$ . Hence $S$ is a vector space over $\mathbb{R}$ .

Prove that $\Phi(a, b) = a + bi$ is an isomorphism $\mathbb{R}^2 \to \mathbb{C}$ of real vector spaces.

\Phi(a, b) = a + bi,\qquad \Psi(a + bi) = (a, b)

\Phi(0, 1) = i

— the imaginary unit

i

sits at the point

(0, 1)

Show proof

Additivity: $\Phi((a,b) + (c,d)) = \Phi(a+c, b+d) = (a+c) + (b+d)i = (a+bi) + (c+di) = \Phi(a,b) + \Phi(c,d)$ .
Homogeneity for real scalars: for real $t$ , $\Phi(t(a,b)) = ta + tbi = t(a+bi) = t\Phi(a,b)$ . The scalars being real is the point — this is an isomorphism of real vector spaces.
Injectivity: $a + bi = c + di$ forces $a = c$ and $b = d$ by uniqueness of real and imaginary parts, so $\Phi$ is injective.
Surjectivity: with $\Psi(a+bi) = (a,b)$ we have $\Psi(\Phi(a,b)) = (a,b)$ and $\Phi(\Psi(a+bi)) = a+bi$ , so $\Psi$ is the inverse of $\Phi$ . A bijective linear map is an isomorphism.

Prove that a linear isomorphism $T: V \to W$ sends linearly independent sets to linearly independent sets.

\sum_{i=1}^{n} c_i\,T(v_i) = 0 \;\Longrightarrow\; c_1 = \cdots = c_n = 0

Sliders

a = 1

b = -1

— the line

y = 1 - x

and the point

(1, -1)

Show proof

Assume $\{v_1, \dots, v_n\} \subset V$ is linearly independent and $\sum c_i T(v_i) = 0$ . By linearity, $\sum c_i T(v_i) = T(\sum c_i v_i)$ , so $T(\sum c_i v_i) = 0$ .
Linear maps send $0$ to $0$ ( $T(0) = T(0 \cdot v) = 0 \cdot T(v) = 0$ ), so injectivity of $T$ gives $\sum c_i v_i = 0$ .
Independence of $\{v_i\}$ forces $c_1 = \cdots = c_n = 0$ . Hence $\{T(v_1), \dots, T(v_n)\}$ is linearly independent.
Reduction to arbitrary (possibly infinite) independent sets: let $E \subseteq V$ be linearly independent. Every finite subset of $T(E)$ has the form $\{T(v_1), \dots, T(v_n)\}$ with distinct $v_i \in E$ (injectivity of $T$ keeps the images distinct), and $\{v_1, \dots, v_n\} \subseteq E$ is independent, so the finite case above makes this finite subset independent.
A set is linearly independent exactly when every finite subset of it is, hence the whole image $T(E)$ is linearly independent.