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?
Best Answer
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.