If $A\in M_{n\times n}^{\mathbb{C}}$ and $T(X)=AX$ for all $X\in M_{n\times \:n}^{\mathbb{C}}$ then $T$ is normal iff $A$ is normal

Question

My try: $\underline{\Rightarrow :}$ Assume $T$ is a normal linear transformation.

Therefore there exists a basis $B=\left\{X_1,...,X_n\right\}$ for $M_{n\times \:n}^{\mathbb{C}}$ that contains eigenvectors of $T$.

Assuming $\lambda _j$ is an egienvalue of the egienvector $X_j$ then

$\lambda _jX_j=T(X_j)=AX_j$

Therefore $\lambda _j$ is an eigenvalue of A and $X_j$ is the corresponding egienvector of the eigenvalue $\lambda _j$ . So we can conclude A has $n$ eigenvectors that form a basis for $M_{n\times \:n}^{\mathbb{C}}$ and therefore A is normal.

the other direction can be showed in a similar manner. In the proof I relied on a theorem that say: If $V$ is a finite vector space and $T:V \to V$ is a linear transformation over $\mathbb{C}$ then $T$ is normal iff there exists an orthonormal basis for $V$ that contains eigenvectors of $T$...

The same goes for A.

I Would like to know what you think about my proof please.

Answer 1

Let us use the scalar product $(X,Y) := \operatorname{Tr}(XY^*)$ on $M_{n\times n}^{\mathbb C}$. Then $$ (T(X),Y) = (AX,Y) = \operatorname{Tr}(AXY^*) = \operatorname{Tr}(XY^*A) = \operatorname{Tr}(X(A^*Y)^*) = (X,A^*Y). $$ Hence, $T^*(Y) = A^*Y$. Now it is really simple to show that $T^*T = TT^*$ is equivalent to $A^*A = AA^*$.