We are working finite-dimensional complex vector space:
If $A$ commutes with $A^*A$, does it follow that $A$ is normal?
Could you please verify my proof:
Since $A^*A$ is positive, let $A^*A = \bigoplus_i^m\lambda_i E_i$ be a spectral form, and let vector space $V = \bigoplus_i^m V_i$ be a direct sum, where $V_i$ is an eigen space of $\lambda_i$.
Liner transformation $A = \bigoplus_i^m A_i$ is reducible by $V_i$. There are multiple ways to prove this. One way is to use the fact that $E_i$ is a projection which is idempotent and which commutes with $A$. Then $A = A\cdot I = A\cdot(\bigoplus_i^m E_i) = A\cdot(\bigoplus_i^m E_i^2) = \bigoplus_i^m E_i A E_i = \bigoplus_i^m A_i$. Another way is to show that each $V_i$ is an invariant under $A$: let $x \in V_i$ such that $A^*Ax = \lambda_i x$ then $A^*A(Ax) = A (A^*Ax) = A\lambda_i x = \lambda_i(Ax)$ and $Ax$ is an eigen vector of $\lambda_i$ and $Ax \in V_i$.
Now that we know that $A$ is reducible, $A^*A = \bigoplus_i^m\lambda_i E_i = \bigoplus_i^m{A_i}^*A_i$ and ${A_i}^*A_i = \lambda_i E_i = \lambda_i I$ on all $V_i$. And if $\lambda_i \neq 0$, $\frac{A_i}{\sqrt \lambda_i}$ is isometry and therefore $A$ is normal. Note: $E_i$ is a projection and it's restriction to $V_i$ is identity $I$. $\lambda_i >= 0$ is positive. Definition of isometry $U^*U = I$. If $\lambda_j = 0$ for some $j$ then $V_j = ker A^*A = ker A$ and $A_j = 0$.