Linear mapping/Matrix to basis/Injective and columns linearly independent/Fact/Proof

Proof

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 with

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

For this vector equation, there exists a nontrivial 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 with ${}\varphi$ not being injective.

To fact