Let $V$ be a finite dimensional inner-product space, and suppose that $T_1$, $T_2$ are normal operators on $V$ that commute. How to show that $T_1+T_2$ and $T_1T_2$ are then normal?
It is clear if $T_1$ commutes with $T_2^*$ and $T_2$ with $T_1^*$, but is it true? If yes how to prove it?
Edit: It is an exercise in Schaum's Outlines of Linear Algebra.
Best Answer
My preferred answer was given by Robert Israel, as it is a great opportunity to meet the Fuglede-Putnam-Rosenblum Theorem.
Here is an interpolation argument in the finite-dimensional case where you stand. I will first prove a lemma for matrices. Of course, the same holds for operators on a finite-dimensional complex inner-product space.
Lemma: a matrix $A\in M_n(\mathbb{C})$ is normal if and only if there exists a polynomial $P\in \mathbb{C}[X]$ such that $A^*=P(A)$.
Proof: the implication $\Leftarrow$ is obvious. So assume $A$ is normal. Equivalently, there exists a unitary matrix $U$ and a diagonal matrix $D=\mbox{Diag}\{\lambda_1,\ldots,\lambda_n\}$ such that $A=UDU^*$. Then $A^*=UD^*U^*$ where $D^*=\mbox{Diag}\{\overline{\lambda_1},\ldots,\overline{\lambda_n}\}$. By Lagrange interpolation, we can find a polynomial $P\in\mathbb{C}[X]$ such that $P(\lambda_j)=\overline{\lambda_j}$ for $j=1,\ldots,n$. Hence $D^*=P(D)$ and it follows that $$P(A)=P(UDU^*)=UP(D)U^*=UD^*U^*=A^*.$$ QED.
Now assume that $T$ is normal and commutes with $S$. Then $S$ commutes with every power of $T$, hence with every polynomial in $T$. By the above, $S$ commutes with $T^*$.
The facts you want to prove follow easily.
Note: this remains true in $\mathbb{R}$. But note that we need to be a little bit careful as real normal matrices are not diagonalizable in $M_n(\mathbb{R})$ in general.