# 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,
$M$
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 $v\in V$,
$\varphi (v)$
$v$
${\mathfrak {v}}$
$M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )$
$m$
${\mathfrak {w}}$

### Proof

$\Box$

## 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

### 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 $\varphi (v_{j})$.
$\Box$
We denote the set of all linear mappings from $V$ to $W$ by $\operatorname {Hom} _{K}{\left(V,W\right)}$. fact means that the mapping
    $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Mat} _{m\times n}(K),\varphi \longmapsto M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ),$
is bijective with the given inverse mapping. A linear mapping
    $\varphi \colon V\longrightarrow V$
is called an endomorphism. The set of all endomorphisms on $V$ is denoted by $\operatorname {End} _{K}{\left(V\right)}$.