Mathematics Lecture Notes

Definition2.5Isomorphism

A linear transformation $T : V \to W$ is an isomorphism if $T$ is bijective (both injective and surjective). We say $V$ and $W$ are isomorphic and write $V \cong W$ .

RemarkInverse is linear

If $T : V \to W$ is an isomorphism, then the inverse function $T^{-1} : W \to V$ is also linear (and hence an isomorphism). For linearity of $T^{-1}$ : if $w_1 = T(v_1)$ and $w_2 = T(v_2)$ , then $T^{-1}(aw_1 + bw_2) = T^{-1}(T(av_1 + bv_2)) = av_1 + bv_2 = aT^{-1}(w_1) + bT^{-1}(w_2)$ .

Theorem2.5Dimension classifies vector spaces

Two finite-dimensional vector spaces over the same field $F$ are isomorphic if and only if they have the same dimension:

$V \cong W \iff \dim V = \dim W.$

In particular, every $n$ -dimensional vector space over $F$ is isomorphic to $F^n$ .

Proof

( $\Rightarrow$ ) If $T : V \to W$ is an isomorphism and $\{v_1, \ldots, v_n\}$ is a basis for $V$ , then $\{T(v_1), \ldots, T(v_n)\}$ is a basis for $W$ (injectivity preserves independence; surjectivity gives spanning). So $\dim W = n = \dim V$ .

( $\Leftarrow$ ) If $\dim V = \dim W = n$ , choose bases $\beta_V = \{v_1, \ldots, v_n\}$ for $V$ and $\beta_W = \{w_1, \ldots, w_n\}$ for $W$ . Define $T(v_i) = w_i$ and extend linearly. Then $T$ is an isomorphism.

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ExampleR^n is isomorphic to P_{n-1}(R)

$\mathbb{R}^n \cong P_{n-1}(\mathbb{R})$ since both have dimension $n$ . An explicit isomorphism: $T(a_1, \ldots, a_n) = a_1 + a_2 x + \cdots + a_n x^{n-1}$ .

ExampleR^4 is isomorphic to M_{2x2}(R)

$\mathbb{R}^4 \cong M_{2 \times 2}(\mathbb{R})$ since both have dimension 4. An explicit isomorphism:

$T(a, b, c, d) = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$

ExampleThe coordinate map

Fix a basis $\beta = \{v_1, \ldots, v_n\}$ for $V$ . The coordinate map $\phi_\beta : V \to F^n$ sending $v = \sum a_i v_i \mapsto (a_1, \ldots, a_n)$ is an isomorphism. This is the canonical isomorphism $V \cong F^n$ .

ExampleSym_2(R) is isomorphic to R^3

$\text{Sym}_2(\mathbb{R}) \cong \mathbb{R}^3$ since $\dim \text{Sym}_2(\mathbb{R}) = 3$ . An explicit isomorphism:

$T\begin{pmatrix} a & b \\ b & c \end{pmatrix} = (a, b, c).$

ExampleR^2 is not isomorphic to R^3

$\mathbb{R}^2 \not\cong \mathbb{R}^3$ because $\dim \mathbb{R}^2 = 2 \neq 3 = \dim \mathbb{R}^3$ .

ExampleC is isomorphic to R^2 (as real vector spaces)

$\mathbb{C} \cong \mathbb{R}^2$ as vector spaces over $\mathbb{R}$ : $T(a + bi) = (a, b)$ . Both have $\dim_{\mathbb{R}} = 2$ .

However, $\mathbb{C} \not\cong \mathbb{R}^2$ as rings (the ring structure of $\mathbb{C}$ has no zero divisors, but this is irrelevant for the vector space isomorphism).

ExampleV is isomorphic to V*

The dual space $V^* = \mathcal{L}(V, F)$ (space of linear functionals) has $\dim V^* = \dim V$ . So $V \cong V^*$ in finite dimensions. However, the isomorphism $V \cong V^*$ is not canonical (it depends on a choice of basis).

The double dual $V^{**} \cong V$ canonically via $v \mapsto \hat{v}$ where $\hat{v}(f) = f(v)$ .

ExampleInfinite-dimensional non-isomorphism

$P(\mathbb{R})$ and $C[0,1]$ are both infinite-dimensional real vector spaces. As real vector spaces, they are not isomorphic: $P(\mathbb{R})$ has a countable basis $\{1, x, x^2, \ldots\}$ , but $C[0,1]$ has no countable basis (its dimension is uncountable).

ExampleAutomorphisms

An isomorphism from $V$ to itself is called an automorphism. The set of all automorphisms of $V$ forms a group under composition, denoted $\text{GL}(V)$ (the general linear group). When $V = F^n$ , this is $\text{GL}_n(F) = \{A \in M_{n \times n}(F) \mid \det A \neq 0\}$ .

ExampleDimension of Hom spaces

$\dim \mathcal{L}(V, W) = \dim V \cdot \dim W$ . If $\dim V = m$ and $\dim W = n$ , then $\mathcal{L}(V, W) \cong M_{n \times m}(F) \cong F^{nm}$ .

ExampleDerivative shift

$D : P_n(\mathbb{R}) \to P_{n-1}(\mathbb{R})$ (differentiation) is surjective but not an isomorphism since $\ker(D) \neq \{0\}$ (constants map to zero). We have $\dim P_n = n + 1 \neq n = \dim P_{n-1}$ .

ExampleIn finite dimensions: injective iff surjective

For a linear map $T : V \to V$ (same finite-dimensional space):

$T$ is injective $\iff$ $T$ is surjective $\iff$ $T$ is an isomorphism.

This follows from $\dim V = \text{rank}(T) + \text{nullity}(T)$ : if $\text{nullity} = 0$ then $\text{rank} = \dim V$ (surjective), and vice versa.

This fails in infinite dimensions: the left shift $L : F^\infty \to F^\infty$ is surjective but not injective.

RemarkEquivalence relation

Isomorphism is an equivalence relation on vector spaces:

Reflexive: $V \cong V$ via $\text{id}_V$ .
Symmetric: If $V \cong W$ via $T$ , then $W \cong V$ via $T^{-1}$ .
Transitive: If $V \cong W$ via $T$ and $W \cong U$ via $S$ , then $V \cong U$ via $S \circ T$ .

The equivalence classes are determined by a single invariant: the dimension.

RemarkLooking ahead

Isomorphic vector spaces are indistinguishable by linear-algebraic properties. Additional structure (inner products, norms, topologies) can distinguish spaces of the same dimension. For instance, all $n$ -dimensional real inner product spaces are isomorphic as inner product spaces, but infinite-dimensional Hilbert spaces are classified by a different notion of dimension.

Math Notes

Isomorphisms of Vector Spaces

Definition

The classification theorem

Examples

Isomorphism as an equivalence relation