Normal transformation Let $V$ be a finite-dimensional vector space over $\Bbb C$ and $T:V\to V$ be a linear transformation. Assume that every eigenvector of $T$ is also an eigenvector of $T^*$.
I need to prove that $TT^*=T^*T$ ($T$ is a normal transformation).
I've managed to show that for all the $V$ subspaces of eigenvectors it is the case.
And if I call the span of all eigenvectors $W$ then the span of non-eigenvectors ($W^\perp$) is orthogonal to it, using that I've shown that $W^\perp$ is an invariant of $T$ and $T^*$ and that's where I'm stuck. It feels like it's close to the end, like there's one last thing missing. Thank you!
 A: Since we are working over a complex inner product space it suffices to show that $V$ has an orthonormal basis consisting of eigenvectors of $T$. Again, since we are working over $\mathbb{C}$, we know that the characteristic polynomial of $T$ splits. Hence by Schur's theorem there exits an orthonormal basis $\{v_1,\dots,v_n\}$ such that the matrix representation of $T$, say $A$, is upper triangular. By definition, we have that $v_1$ is an eigenvector of $T$. We prove that each $v_i$ is also an eigenvector by induction. Suppose that $v_1,\dots,v_{k-1}$ are eigenvectors for some $k$. For $j<k$, let $\lambda_j$ be the eigenvector corresponding to $v_j$. Since $A$ is upper triangular $$T(v_k)=A_{1k}v_1+A_{2k}v_2+\dots+A_{kk}v_k.$$ Since our basis is orthonormal we have that $$A_{jk}=\langle T(v_k),v_j\rangle=\langle v_k, T^\ast(v_j)\rangle=\langle v_k,\eta_j v_j\rangle=\overline{\eta_j}\langle v_k,v_j\rangle=0$$where $\eta_j$ is the eigenvalue of $T^\ast$ corresponding to $v_j$. Since this holds for all $j<k$ it must be that $T(v_k)=A_{kk}v_k$.
