4 Linear algebra 4.21 Change of basis Further reading

4.22 Isomorphisms

Consider the real vector space of all height 3 column vectors, and the real vector space of all height 3 row vectors. These are different vector spaces but they are essentially the same: there’s no vector space property which one has but the other doesn’t.

A way to make this precise is to observe that there is a bijective linear map between them, namely the transpose which sends $\begin{pmatrix}x\\ y\\ z\end{pmatrix}$ to $(x\;y\;z)$ .

Definition 4.22.1.

•

Linear maps which are bijections are called vector space isomorphisms, or just isomorphisms.
•

If there is an isomorphism $U\to V$ , we say that $U$ and $V$ are isomorphic and write $U\cong V$ .

Notice that the existence of a linear map between two vector spaces requires that they have the same field of scalars.

Lemma 4.22.1.

A linear map $T:U\to V$ is one-to-one if and only if $\ker T=\{\mathbf{0}_{U}\}$ .

Proof.

: Suppose $T$ is one-to-one. We know $T(\mathbf{0}_{U})=\mathbf{0}_{V}$ . Suppose $\mathbf{x}\in\ker T$ . Then $T(\mathbf{x})=\mathbf{0}_{V}=T(\mathbf{0}_{U})$ , so $\mathbf{x}=\mathbf{0}_{U}$ . Thus $\ker T=\{\mathbf{0}_{U}\}$ .

Suppose $\ker T=\{\mathbf{0}_{U}\}$ and that $T(\mathbf{x})=T(\mathbf{y})$ . Then $T(\mathbf{x}-\mathbf{y})=\mathbf{0}_{V}$ by linearity, so $\mathbf{x}-\mathbf{y}\in\ker T$ , so $\mathbf{x}-\mathbf{y}=0_{U}$ , so $\mathbf{x}=\mathbf{y}$ and $T$ is one-to-one. ∎

Lemma 4.22.2.

If $U\cong V$ then $\dim U=\dim V$ .

Proof.

There’s a linear bijection $T:U\to V$ . $\ker T=\{\mathbf{0}_{U}\}$ , so applying the rank-nullity theorem $\dim U=\dim\ker T+\dim\operatorname{im}T=\dim\operatorname{im}T$ . Since $\ker T=\{\mathbf{0}_{U}\}$ we get $\dim U=\dim\operatorname{im}T$ , but $T$ is onto so $\operatorname{im}T=V$ and $\dim U=\dim V$ . ∎

Perhaps more surprisingly, the converse is true. If two vector spaces over the same field have the same dimension, they’re isomorphic. The first step to proving this is:

Proposition 4.22.3.

Let $V$ be an $\mathbb{F}$ -vector space with dimension $n$ . Then $\mathbb{F}^{n}\cong V$ .

Proof.

Let $\mathbf{b}_{1},\ldots,\mathbf{b}_{n}$ be a basis of $V$ , and define a map $T:\mathbb{F}^{n}\to V$ by

T\begin{pmatrix}x_{1}\\ \vdots\\ x_{n}\end{pmatrix}=\sum_{i=1}^{n}x_{i}\mathbf{b}_{i}.

We have to show that $T$ is linear, one-to-one, and onto.

$T$ is linear because

	$\displaystyle T\left(\begin{pmatrix}x_{1}\\ \vdots\\ x_{n}\end{pmatrix}+\begin{pmatrix}y_{1}\\ \vdots\\ y_{n}\end{pmatrix}\right)$	$\displaystyle=T\begin{pmatrix}x_{1}+y_{1}\\ \vdots\\ x_{n}+y_{n}\end{pmatrix}$
		$\displaystyle=\sum_{i=1}^{n}(x_{i}+y_{i})\mathbf{b}_{i}$
		$\displaystyle=\sum_{i=1}^{n}x_{i}\mathbf{b}_{i}+\sum_{i=1}^{n}y_{i}\mathbf{b}_% {i}$
		$\displaystyle=T\begin{pmatrix}x_{1}\\ \vdots\\ x_{n}\end{pmatrix}+T\begin{pmatrix}y_{1}\\ \vdots\\ y_{n}\end{pmatrix}$

and

	$\displaystyle T\left(\lambda\begin{pmatrix}x_{1}\\ \vdots\\ x_{n}\end{pmatrix}\right)$	$\displaystyle=T\begin{pmatrix}\lambda x_{1}\\ \vdots\\ \lambda x_{n}\end{pmatrix}$
		$\displaystyle=\sum_{i=1}^{n}(\lambda x_{i})\mathbf{b}_{i}$
		$\displaystyle=\lambda\sum_{i=1}^{n}x_{i}\mathbf{b}_{i}$
		$\displaystyle=\lambda T\begin{pmatrix}x_{1}\\ \vdots\\ x_{n}\end{pmatrix}.$

$T$ is one-to-one because if $\sum_{i=1}x_{i}\mathbf{b}_{i}=\sum_{i=1}^{n}y_{i}\mathbf{b}_{i}$ then $x_{i}=y_{i}$ for all $i$ by Lemma 4.8.1.

$T$ is onto because the image of $T$ is the set of linear combinations of $\mathbf{b}_{1},\ldots,\mathbf{b}_{n}$ , which is all of $V$ because $\mathbf{b}_{1},\ldots,\mathbf{b}_{n}$ is a spanning sequence. ∎

Now given any two $\mathbb{F}$ -vector spaces $U$ and $V$ with dimension $n$ we know there is an isomorphism $f:\mathbb{F}^{n}\to U$ and another isomorphism $g:\mathbb{F}^{n}\to V$ . We’d like to get an isomorphism from $U$ to $V$ and it seems like the right thing is $g\circ f^{-1}$ . We know that the inverse of a bijection is a bijection, so $f^{-1}$ is a bijection, but we don’t know that it is linear.

Lemma 4.22.4.

Let $X$ and $Y$ be $\mathbb{F}$ -vector spaces and $T:X\to Y$ be a linear bijection. Then $T^{-1}:Y\to X$ is linear.

Proof.

We check the two parts of the definition of linearity.

1.

Let $\mathbf{y}\in Y$ and $\lambda\in\mathbb{F}$ . Since $T$ is onto, $\mathbf{y}=T(\mathbf{x})$ for some $\mathbf{x}\in X$ , and $T^{-1}(\mathbf{y})=\mathbf{x}$ . Now multiplying $\mathbf{y}=T(\mathbf{x})$ by $\lambda$ give $\lambda\mathbf{y}=\lambda T(\mathbf{x})=T(\lambda\mathbf{x})$ since $T$ is linear, and so $T^{-1}(\lambda\mathbf{y})=\lambda\mathbf{x}=\lambda T^{-1}(\mathbf{y})$ .
2.

Let $\mathbf{y}_{1},\mathbf{y}_{2}\in Y$ . Again, as $T$ is onto, there exist $\mathbf{x}_{1},\mathbf{x}_{2}\in X$ such that $T(\mathbf{x}_{i})=\mathbf{y}_{i}$ , and consequently $T^{-1}(\mathbf{y}_{i})=\mathbf{x}_{i}$ . Now $\mathbf{y}_{1}+\mathbf{y}_{2}=T(\mathbf{x}_{1})+T(\mathbf{x}_{2})=T(\mathbf{x}% _{1}+\mathbf{x}_{2})$ as $T$ is linear, so $T^{-1}(\mathbf{y}_{1}+\mathbf{y}_{2})=\mathbf{x}_{1}+\mathbf{x}_{2}=T^{-1}(% \mathbf{y}_{1})+T^{-1}(\mathbf{y}_{2})$ . ∎

Theorem 4.22.5.

Any two vector spaces of the same finite dimension over the same field $\mathbb{F}$ are isomorphic.

Proof.

We have already seen that if $\dim U=\dim V=n$ then there are isomorphisms $f:\mathbb{F}^{n}\to U$ and $g:\mathbb{F}^{n}\to V$ . The map $g\circ f^{-1}$ is a bijection because it is the composition of two bijections, and is linear because by the previous lemma it is the composition of two linear maps. Therefore it is a vector space isomorphism, and $U\cong V$ . ∎