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? 
 A: Suppose that $T$ is bijective. Then $T$ is surjective, and so, there exists a matrix $B$ such that $T(B)=I$, and so $AB=I$. Youcan show that this means $A$ is invertible.
Similarly if $A$ is inverible, then $T(A^{-1}B)=B$ for any matrix $B$ in $M_n(K)$. Hence $T$ is surjective. It's easy to show $T$ is injective as well.
A: 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)
A: Hint: Note that a function is bijective if and only if it has a two-sided inverse. Given that $A$ has an inverse $A^{-1}$ such that $A^{-1}A=AA^{-1}=\mathbb 1$, can you come up with an two-sided inverse $T^{-1}: M_n(K) \rightarrow M_n(K)$ for $T$? For the converse, consider $T^{-1}(A)$.
