Here is a question as phrased in Advanced Linear Algebra (Steven Roman):
Show that a linear operator $\tau$ on a finite-dimensional complex inner product space $V$ is normal if and only if whenever $S$ is an invariant subspace under $\tau$, so is $S^\perp$.
I'm only interested in the $\Rightarrow$ direction. Wikipedia (and ProofWiki) has a proof that uses some kind of computation with the trace. Linear Algebra Done Right chooses an orthonormal basis and then uses the fact that $\Vert \tau v \Vert = \Vert \tau^* v \Vert$.
However, the above proofs seem a bit too computational to me. Can the following argument work?
(Use $\odot$ for orthogonal direct sum.) Suppose that $\tau$ is normal and write $$V=E_{\lambda_1} \odot \cdots \odot E_{\lambda_k}$$ where the $E_{\lambda_i}$ are the eigenspaces of $\tau$. If $S$ is a $\tau$-invariant subspace then $\tau|_S$ is diagonalizable, so $$S=\widetilde{E}_{\lambda_1} \odot\cdots\odot \widetilde{E}_{\lambda_k}$$ where the $\widetilde{E}_{\lambda_i} \le E_{\lambda_i}$ are the eigenspaces of $\tau|_S$ (which could be trivial). Then choose orthogonal complements $D_i$ such that $E_{\lambda_i} = \widetilde{E}_{\lambda_i} \odot D_i$, and then $$S^\perp = D_1 \odot\cdots\odot D_k.$$ Since each $D_i$ is a subspace of an eigenspace, they are all $\tau$-invariant.
Best Answer
This proof seems to be correct. The only problem with it, is that you already use that $\tau$ is diagonalizable (when you wrote $V$ as the orthogonal direct sum of eigenspaces), but to prove that, one usually uses this lemma...