Show that a linear matrix transformation is bijective iff A is invertible. Suppose a linear transformation $T: M_n(K) \rightarrow M_n(K)$ defined by $T(M) = A M$ for $M \in M_n(K)$.
Show that it is bijective IFF $A$ is invertible. 
I was thinking then that I could show that it is surjective. So suppose there exists a 
$B \in  M_n(K)$ such that $T(B) = ?$ What would it equal to show that? 

By the rank-nullity theorem $T$ is bijective if and only if $T$ is injective if and only if $T$ is surjective.
For the injectivity: we have
$$T(M)=T(N)\iff AM=AN$$


*

*If $A$ is invertible then $AM=AN\implies M=N$ and then $T$ is injective.

*If $A$ isn't invertible so let $N=M+(x\; x\;\cdots\; x)$ where $x\in\ker A, x\ne0$ and we have $N\ne M$ but  $T(M)=T(N)$ so $T$ isn't injective.(Proof by contapositive)
