How can I prove that the B-matrix of T (a linear transformation) is injective/surjective implies that T is injective/surjective?

Question

T is a linear transformation and B-matrix of T with respect to any arbitrary basis.

This seems intuitively true to me because changing the transformation with respect to a new basis shouldn't change anything, but I just want to confirm this. Also, if you can illustrate a proof, that would be great.

Answer 1

A linear transformation is completely determined by the image of a basis that is precisely by its matrix representations in that basis.

Refer also to the related determining a linear transformation.

Answer 2

Let $\varphi$ be the canonical coordinate change with respect to $B$, so $T=\varphi^{-1}\circ B\circ\varphi$, if $B$ is injective and $Tx=0$, then $\varphi^{-1}(B\varphi (x))=0$, and hence $B\varphi(x)=0$, then $\varphi(x)=0$, and finally $x=0$.

Can you prove the surjectivity?