Linear mapping/Matrix to basis/Several properties/Fact/Proof

Proof

Let ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$ denote the bases of ${}V$ and ${}W$, respectively. Let ${}s_{1},\ldots ,s_{n}$ denote the column vectors of ${}M$. (1). The mapping ${}\varphi$ has the property

${}\varphi (v_{j})=\sum _{i=1}^{m}s_{ij}w_{i}\,,$

where ${}s_{ij}$ is the ${}i$-th entry of the ${}j$-th column vector ${}s_{j}$. Therefore,

${}\varphi {\left(\sum _{j=1}^{n}a_{j}v_{j}\right)}=\sum _{j=1}^{n}a_{j}{\left(\sum _{i=1}^{m}s_{ij}w_{i}\right)}=\sum _{i=1}^{m}{\left(\sum _{j=1}^{n}a_{j}s_{ij}\right)}w_{i}\,.$

This is ${}0$ if and only if ${}\sum _{j=1}^{n}a_{j}s_{ij}=0$ for all ${}i$, and this is equivalent to

${}\sum _{j=1}^{n}a_{j}s_{j}=0\,.$

For this vector equation, there exists a nontrivial solution tuple ${}{\left(a_{1},\ldots ,a_{n}\right)}$ if and only if the columns are linearly dependent, and this holds if and only if the kernel of ${}\varphi$ is not trivial. Due to fact, this is equivalent to ${}\varphi$ not being injective.
(2). See exercise.
(3). Let ${}n=m$. The first equivalence follows from (1) and (2). If ${}\varphi$ is bijective, then there exists a (linear) inverse mapping ${}\varphi ^{-1}$ with

$\varphi \circ \varphi ^{-1}=\operatorname {Id} _{W}{\text{ and }}\varphi ^{-1}\circ \varphi =\operatorname {Id} _{V}.$

Let ${}M$ denote the matrix for ${}\varphi$, and ${}N$ the matrix for ${}\varphi ^{-1}$. The matrix for the identity is the identity matrix. Because of fact, we have

${}M\circ N=E_{n}=N\circ M\,;$

therefore, ${}M$ is invertible. The reverse implication is proved similarly.

To fact