If $A$ is normal such that $AB=BA$, then $A^*B=BA^*$

Question

Please help me.

Let $A,B\in M_n(\mathbb{C})$. Show that if $A$ is normal such that $AB=BA$, then $A^*B=BA^*$.

Answer 1

As $A$ is normal, it is orthogonally diagonalizable. So, there is an orthogonal matrix $P$ such that: $$A=P^*DP,$$ where $D$ is a diagonal matrix. Clearly, $A^*=P^*D^*P$.

Then there is a polynomial $p=p(x)=\sum_{k=0}^mc_kx^k$, such that: $p(D)=D^*$.

Indeed, if $D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$, then $p(D)=\mathrm{diag}\big(p(\lambda_1),\ldots,p(\lambda_n)\big)$, and this polynomial is chosen so that it takes $\lambda_j$ to $\overline{\lambda}_j$. But $$ p(A)=\sum_{k=0}^m c_kA^k=\sum_{k=0}^m c_k(P^*DP)^k=\sum_{k=0}^m c_kP^*D^kP=P^*\left(\sum_{k=0}^m c_kD^k\right)P\\=P^*p(D)P=P^*D^*P=A^*. $$ Since $AB=BA$, then $A^kB=BA^k$, and hence $p(A)B=Bp(A)$. Thus, $$A^*B=p(A)B=Bp(A)=BA^*.$$