Linear mapping/Matrix for basis/Surjective and column generating system/Exercise/Solution

We consider the commutative diagramm

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

Since the coordinate mappings are bijective, the map ${}\varphi$ is surjective if and only if ${}M=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )$ is surjective. The image vector of the ${}j$th standard vector ${}e_{j}$ under ${}M$ is the ${}j$th column of ${}M$ and the image space for ${}M$ is the space generated by the columnns. Therefore surjectivity is equivalent with the property that the columns form a generating system of ${}K^{m}$.

To the exercise