I tried to solve the following question to no avail and will appreciate any hints.
Let $M \in M_{n×n}(C)$ be an invertible matrix. We'll define the transformation $S : M_{n×n}(C) \to M_{n×n}(C)$ by the following rule, for any matrix $X \in M_{n×n}(C)$ , $S(X) = XM^{-1}$. Prove that if $M$ is normal then $S$ is normal.
I tried to show that $M^{-1}$ and the transformation $S$ have the same minimal polynomial and because $M^{-1}$ is normal, and unitary diagonalizable then $S$ is diagonalizable but I have no idea how to show that it is unitary diagonalizable.
Editing my question: I tried also to do the following (which is using the way of how we define the inner product between two elements in $M_{n×n}(C)$:
$<S(X),Y> = trace(Y^{*}XM^{-1}) = <X,S^*(Y)> \\ $
But have no idea how to conclude what $S^*(X)$ is.
I will appreciate any clues, thanks !
Use conjugate symmetry of the inner product to write \begin{align*} \overline{\langle S(X),Y\rangle} &= \langle Y,S(X)\rangle = \text{tr}(S(X)^*Y) = \text{tr}((XM^{-1})^*Y) \\ &= \text{tr}((M^{-1})^*X^*Y) = \text{tr}(X^*Y(M^{-1})^*) \\ &= \langle Y(M^{-1})^*,X\rangle = \overline{\langle X,Y(M^{-1})^*\rangle}. \end{align*} Note that the essential step, on the second line, is the fact that $\text{tr}(AB) = \text{tr}(BA)$. From this we conclude (since $X,Y$ are arbitrary) that $S^*(Y) = Y(M^{-1})^*$.
Now I leave it to you to verify that $S(S^*(X)) = S^*(S(X))$.