4 Linear algebra 4.20 Change of basis Further reading

4.21 Coordinate isomorphisms

4.21.1 Vector space 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 4.21.1.

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Linear maps which are bijections are called vector space isomorphisms, or just isomorphisms.
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If there is an isomorphism $U\to V$ , we say that $U$ and $V$ are isomorphic and write $U\cong V$ .

Isomorphic vector spaces share the same vector space properties — for example, they always have the same dimension.

Lemma 4.21.1.

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 $\dim U=\dim V$ . ∎

4.21.2 Coordinate isomorphisms

If $V$ is a finite-dimensional $\mathbb{F}$ -vector space with a basis $\mathcal{B}=\mathbf{b}_{1},\ldots,\mathbf{b}_{n}$ then every $\mathbf{v}\in V$ can be written uniquely as $a_{1}\mathbf{b}_{1}+\cdots+a_{n}\mathbf{b}_{n}$ for some $a_{i}\in\mathbb{F}$ . The column vector

[\mathbf{v}]_{\mathcal{B}}=\begin{pmatrix}a_{1}\\ \vdots\\ a_{n}\end{pmatrix}

is called the coordinate vector of $\mathbf{v}$ with respect to $\mathcal{B}$ . We can use this idea to define an isomorphism between $V$ and $\mathbb{F}^{n}$ .

Proposition 4.21.2.

Let $V$ be a $\mathbb{F}$ -vector space with basis $\mathcal{B}=\mathbf{b}_{1},\ldots,\mathbf{b}_{n}$ . Define $[-]_{\mathcal{B}}:V\to\mathbb{F}^{n}$ to be the function that sends $\mathbf{v}\in V$ to its coordinate vector $[\mathbf{v}]_{\mathcal{B}}$ . Then $[-]_{\mathcal{B}}$ is a vector space isomorphism.

$[-]_{\mathcal{B}}$ is called the coordinate isomorphism with respect to $\mathcal{B}$ .

Proof.

First we have to show it is linear.

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Let $\mathbf{x},\mathbf{y}\in V$ and let $\mathbf{x}=\sum_{i}x_{i}\mathbf{b}_{i}$ and $\mathbf{y}=\sum_{i}y_{i}\mathbf{b}_{i}$ , so that $[\mathbf{x}]_{\mathcal{B}}=(x_{1},\ldots,x_{n})^{T}$ and $[\mathbf{y}]=(y_{1},\ldots,y_{n})^{T}$ . Then $\mathbf{x}+\mathbf{y}=\sum_{i}(x_{i}+y_{i})\mathbf{b}_{i}$ , so

$\displaystyle{[}\mathbf{x}+\mathbf{y}]_{\mathcal{B}}$ $\displaystyle=(x_{1}+y_{1},\ldots,x_{n}+y_{n})^{T}$

$\displaystyle=(x_{1},\ldots,x_{n})^{T}+(y_{1},\ldots,y_{n})^{T}$

$\displaystyle=[\mathbf{x}]_{\mathcal{B}}+[\mathbf{y}]_{\mathcal{B}}.$
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Let $\mathbf{x}\in V$ and $\lambda\in\mathbb{F}$ . If $\mathbf{x}=\sum_{i}x_{i}\mathbf{b}_{i}$ so that $[\mathbf{x}]_{\mathcal{B}}=(x_{1},\ldots,x_{n})^{T}$ , then $\lambda\mathbf{x}=\sum_{i}\lambda x_{i}\mathbf{b}_{i}$ so $[\lambda\mathbf{x}]_{\mathcal{B}}=(\lambda x_{1},\ldots,\lambda x_{n})^{T}=% \lambda(x_{1},\ldots,x_{n})^{T}=\lambda[\mathbf{x}]_{\mathcal{B}}$ .

Now we have to show it is injective and surjective.

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If $[\mathbf{v}]_{\mathcal{B}}=\mathbf{0}_{n}$ then by definition $\mathbf{v}=0\mathbf{b}_{1}+\cdots+0\mathbf{b}_{n}=\mathbf{0}_{V}$ . Therefore $\ker[-]_{\mathcal{B}}=\{\mathbf{0}_{V}\}$ and by Proposition 4.14.3, $[-]_{\mathcal{B}}$ is injective.
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Any vector $(x_{1},\ldots,x_{n})^{T}$ in $\mathbb{F}^{n}$ is the image of $\sum_{i}x_{i}\mathbf{b}_{i}\in V$ under $[-]_{\mathcal{B}}$ .

∎

In particular, every finite-dimensional vector space is isomorphic to a space of column vectors.

4.21.3 Coordinate and matrices of linear maps

Suppose we have $\mathbb{F}$ -vector spaces $U, V$ of dimensions $n$ and $m$ with bases $\mathcal{B}=\mathbf{b}_{1},\ldots,\mathbf{b}_{n},\mathcal{C}=\mathbf{c}_{1},% \ldots,\mathbf{c}_{m}$ and a linear map $T:U\to V$ . We have coordinate isomorphisms $[-]_{\mathcal{B}}:U\to\mathbb{F}^{n}$ and $[-]_{\mathcal{C}}:V\to\mathbb{F}^{m}$ and a matrix $[T]^{\mathcal{B}}_{\mathcal{C}}\in M_{m\times n}(\mathbb{F})$ . What is the relationship between $T:U\to V$ and the linear map $\mathbb{F}^{n}\to\mathbb{F}^{m}$ given by left-multiplication by $[T]^{\mathcal{B}}_{\mathcal{C}}$ ?

Theorem 4.21.3.

In the notation above, for any $\mathbf{u}\in U$ we have $[T]^{\mathcal{B}}_{\mathcal{C}}[\mathbf{u}]_{\mathcal{B}}=[T(\mathbf{u})]_{% \mathcal{C}}$ .

Proof.

Let $[T]^{\mathcal{B}}_{\mathcal{C}}=(a_{ij})$ , so that $T(\mathbf{b}_{j})=\sum_{i=1}^{m}a_{ij}\mathbf{c}_{j}$ . Notice that this says $[T(\mathbf{b}_{j})]_{\mathcal{C}}$ is $(a_{1j},\ldots,a_{mj})^{T}$ , the $j$ th column of $[T]^{\mathcal{B}}_{\mathcal{C}}$ . The coordinate vector $[\mathbf{b}_{j}]_{\mathcal{B}}$ is the $j$ th standard basis vector $\mathbf{e}_{j}$ , so $[T]^{\mathcal{B}}_{\mathcal{C}}[\mathbf{b}_{j}]=[T]^{\mathcal{B}}_{\mathcal{C}% }\mathbf{e}_{j}$ which equals the $j$ th column of $[T]^{\mathcal{B}}_{\mathcal{C}}$ . Therefore

[T]^{\mathcal{B}}_{\mathcal{C}}[\mathbf{b}_{j}]_{\mathcal{B}}=[T(\mathbf{b}_{j% })]_{\mathcal{C}}.

(4.8)

Let $\mathbf{u}\in U$ and write $\mathbf{u}=\sum_{j}u_{j}\mathbf{b}_{j}$ so that $[\mathbf{u}]_{\mathcal{B}}=(u_{1},\ldots,u_{n})^{T}$ . Then

$\displaystyle[T(\mathbf{u})]_{\mathcal{C}}$	$\displaystyle=\left[T\left(\sum_{j}u_{j}\mathbf{b}_{j}\right)\right]$
	$\displaystyle=\left[\sum_{j}u_{j}T(\mathbf{b}_{j})\right]_{\mathcal{C}}$	$\displaystyle\text{linearity of }T$
	$\displaystyle=\sum_{j}u_{j}[T(\mathbf{b}_{j})]_{\mathcal{C}}$	$\displaystyle\text{linearity of }[-]_{\mathcal{C}}$
	$\displaystyle=\sum_{j}u_{j}[T]^{\mathcal{B}}_{\mathcal{C}}[\mathbf{b}_{j}]_{% \mathcal{B}}$	(4.8)
	$\displaystyle=[T]^{\mathcal{B}}_{\mathcal{C}}\sum_{j}u_{j}[\mathbf{b}_{j}]$
	$\displaystyle=[T]^{\mathcal{B}}_{\mathcal{C}}\left[\sum_{j}u_{j}\mathbf{b}_{j}% \right]_{\mathcal{B}}$	$\displaystyle\text{linearity of }[-]_{\mathcal{B}}$
	$\displaystyle=[T]^{\mathcal{B}}_{\mathcal{C}}[\mathbf{u}]_{\mathcal{B}}.$

∎

We can now answer the following natural question: if $\mathcal{B}$ and $\mathcal{C}$ are two bases of the finite-dimensional vector space $V$ and $\mathbf{v}\in V$ , what is the relationship between $[\mathbf{v}]_{\mathcal{B}}$ and $[\mathbf{v}]_{\mathcal{C}}$ ? Using the previous theorem applied to the identity map $\operatorname{id}:V\to V$ ,

	$\displaystyle[\mathbf{v}]_{\mathcal{C}}$	$\displaystyle=[\operatorname{id}(\mathbf{v})]_{\mathcal{C}}$
		$\displaystyle=[\operatorname{id}]^{\mathcal{B}}_{\mathcal{C}}[\mathbf{v}]_{% \mathcal{B}}.$

	$\displaystyle{[}\mathbf{x}+\mathbf{y}]_{\mathcal{B}}$	$\displaystyle=(x_{1}+y_{1},\ldots,x_{n}+y_{n})^{T}$
		$\displaystyle=(x_{1},\ldots,x_{n})^{T}+(y_{1},\ldots,y_{n})^{T}$
		$\displaystyle=[\mathbf{x}]_{\mathcal{B}}+[\mathbf{y}]_{\mathcal{B}}.$