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?

  • $\begingroup$ What is the set $M_n(K)$? $\endgroup$ – mavavilj Jan 17 '16 at 16:17

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.


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)
  • $\begingroup$ What are $N$ and $M$? $\endgroup$ – mavavilj Jan 17 '16 at 16:16

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 $T^{-1}(A)$.


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