## The matrix of a nonsingular linear transformation is nonsingular, regardless of the bases chosen

Let $F$ be a field, and let $V$ and $W$ be $n$– and $m$-dimensional vector spaces over $F$, respectively. Prove that, for any bases $B \subseteq V$ and $E \subseteq W$, $\varphi : V \rightarrow W$ is a nonsingular linear transformation if and only if $M^E_B(\varphi)$ is nonsingular matrix.

Choose some basis $B \subseteq V$ and $E \subseteq W$.

Suppose $\varphi : V \rightarrow W$ is a nonsingular linear transformation. Then $\mathsf{ker}\ \varphi = 0$. Now suppose $x \in V$. If $M^E_B(\varphi) \times x = 0$, then $\varphi(x) = 0$, and so $x = 0$. Thus $M^E_B(\varphi)$ is a nonsingular matrix.

Conversely, suppose $M^E_B(\varphi)$ is a nonsingular matrix. Suppose $x \in \mathsf{ker}\ \varphi$. Then $\varphi(x) = 0$, and so we have $M^E_B(\varphi) \times x = 0$. Since $M^E_B(\varphi)$ is nonsingular, $x = 0$. Thus $\mathsf{ker}\ \varphi = 0$, and so $\varphi$ is nonsingular as a linear transformation.