Linear mapping/Matrix/Relation/2/Section

A linear mapping

\varphi \colon K^{n}\longrightarrow K^{m}

is determined uniquely by the images ${}\varphi (e_{j})$ , ${}j=1,\ldots ,n$ , of the standard vectors, and every ${}\varphi (e_{j})$ is a linear combination

{}\varphi (e_{j})=\sum _{i=1}^{m}a_{ij}e_{i}\,

and hence determined by the elements ${}a_{ij}$ . This means all together that such a linear mapping is given by the ${}mn$ elements ${}a_{ij}$ , ${}1\leq i\leq m$ , ${}1\leq j\leq n$ . Such a set of data can be written as a matrix. Due to fact, this observation holds for all finite-dimensional vector spaces, as long as bases are fixed on the source space and on the target space of the linear mapping.

Definition

Let ${}K$ denote a field, and let ${}V$ be an ${}n$ -dimensional vector space with a basis ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ , and let ${}W$ be an ${}m$ -dimensional vector space with a basis ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$ .

For a linear mapping

\varphi \colon V\longrightarrow W,

the matrix

{}M=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )=(a_{ij})_{ij}\,,

where ${}a_{ij}$ is the ${}i$ -th coordinate of ${}\varphi (v_{j})$ with respect to the basis ${}{\mathfrak {w}}$ , is called the describing matrix for ${}\varphi$ with respect to the bases.

For a matrix ${}M=(a_{ij})_{ij}\in \operatorname {Mat} _{m\times n}(K)$ , the linear mapping ${}\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)$ determined by

v_{j}\longmapsto \sum _{i=1}^{m}a_{ij}w_{i}

in the sense of fact,

is called the linear mapping determined by the matrix

{}M

.

For a linear mapping ${}\varphi \colon K^{n}\rightarrow K^{m}$ , we always assume that everything is with respect to the standard bases, unless otherwise stated. For a linear mapping ${}\varphi \colon V\rightarrow V$ from a vector space in itself (what is called an endomorphism), one usually takes the same bases on both sides. The identity on a vector space of dimension ${}n$ is described by the identity matrix, with respect to every basis.

If ${}V=W$ , then we are usually interested in the describing matrix with respect to one basis ${}{\mathfrak {v}}$ of ${}V$ .

Example

Let ${}V$ denote a vector space with bases ${}{\mathfrak {v}}$ and ${}{\mathfrak {w}}$ . If we consider the identity

\operatorname {Id} \colon V\longrightarrow V

with respect to the basis ${}{\mathfrak {v}}$ on the source and the basis ${}{\mathfrak {w}}$ on the target, we get, because of

{}\operatorname {Id} (v_{j})=v_{j}=\sum a_{ij}w_{i}\,,

directly

{}M_{\mathfrak {w}}^{\mathfrak {v}}(\operatorname {Id} )=M_{\mathfrak {w}}^{\mathfrak {v}}\,.

This means that the describing matrix of the identical linear mapping is the transformation matrix for the base change from ${}{\mathfrak {v}}$ to ${}{\mathfrak {w}}$ .

Lemma

Let ${}K$ denote a field and let ${}V$ denote an ${}n$ -dimensional vector space with a basis ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ . Let ${}W$ be an ${}m$ -dimensional vector space with a basis ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$ , and let

\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V

and

\Psi _{\mathfrak {w}}\colon K^{m}\longrightarrow W

be the corresponding mappings. Let

\varphi \colon V\longrightarrow W

denote a linear mapping with describing matrix ${}M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )$ . Then

{}\varphi \circ \Psi _{\mathfrak {v}}=\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\,

hold, that is, the diagram

{\begin{matrix}K^{n}&{\stackrel {\Psi _{\mathfrak {v}}}{\longrightarrow }}&V&\\\!\!\!\!\!M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\downarrow &&\downarrow \varphi \!\!\!\!\!&\\K^{m}&{\stackrel {\Psi _{\mathfrak {w}}}{\longrightarrow }}&W&\!\!\!\!\!\\\end{matrix}}

commutes. For a vector ${}v\in V$ ,

we can compute

{}\varphi (v)

by determining the coefficient tuple of

{}v

with respect to the basis

{}{\mathfrak {v}}

, applying the matrix

{}M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )

and determining for the resulting

{}m

-tuple the corresponding vector with respect to

{}{\mathfrak {w}}

.

Proof

See exercise.

\Box

Theorem

Let ${}K$ be a field, and let ${}V$ be an ${}n$ -dimensional vector space with a basis ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ , and let ${}W$ be an ${}m$ -dimensional vector space with a basis ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$ . Then the mappings

\varphi \longmapsto M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ){\text{ and }}M\longmapsto \varphi _{\mathfrak {w}}^{\mathfrak {v}}(M),

defined in definition, are inverse

to each other.

Proof

We show that both compositions are the identity. We start with a matrix ${}M={\left(a_{ij}\right)}_{ij}$ and consider the matrix

M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)).

Two matrices are equal, when the entries coincide for every index pair ${}(i,j)$ . We have

{}{\begin{aligned}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)))_{ij}&=i-{\text{th coordinate of }}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M))(v_{j})\\&=i-{\text{th coordinate of }}\sum _{i=1}^{m}a_{ij}w_{i}\\&=a_{ij}.\end{aligned}}

Now, let ${}\varphi$ be a linear mapping, we consider

\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )).

Two linear mappings coincide, due to fact, when they have the same values on the basis ${}v_{1},\ldots ,v_{n}$ . We have

{}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )))(v_{j})=\sum _{i=1}^{m}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))_{ij}\,w_{i}\,.

Due to the definition, the coefficient ${}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))_{ij}$ is the ${}i$ -th coordinate of ${}\varphi (v_{j})$ with respect to the basis ${}w_{1},\ldots ,w_{m}$ . Hence, this sum equals ${}\varphi (v_{j})$ .