Making a basis out of the eigenvalues of a normal endomorphism without using the Spectral Theorem

Question

Consider the following question:

Define the adjoint $\alpha^*$ of an endomorphism $\alpha$ of a complex inner-product space $V$. Show that if $W$ is a subspace of $V$, then $\alpha(W) \subseteq W$ if and only if $\alpha^*\left(W^{\perp}\right) > \subseteq W^{\perp}$.

An endomorphism of a complex inner-product space is said to be normal if it commutes with its adjoint. Prove the following facts about a normal endomorphism $\alpha$ of a finite-dimensional space $V$.

(i) $\alpha$ and $\alpha^*$ have the same kernel.

(ii) $\alpha$ and $\alpha^*$ have the same eigenvectors, with complex conjugate eigenvalues.

(iii) If $E_\lambda=\{x \in V: \alpha(x)=\lambda x\}$, then $\alpha\left(E_\lambda^{\perp}\right) \subseteq E_\lambda^{\perp}$.

(iv) There is an orthonormal basis of $V$ consisting of eigenvectors of $\alpha$.

Deduce that an endomorphism $\alpha$ is normal if and only if it can be written as a product $\beta \gamma$, where $\beta$ is Hermitian, $\gamma$ is unitary and $\beta$ and $\gamma$ commute with each other. [Hint: Given $\alpha$, define $\beta$ and $\gamma$ in terms of their effect on the basis constructed in (iv).]

I have already seen a proof of (iv) with the spectral theorem, however, in this course the theorem was not lectured. (I checked similar questions here but they all seem to be using it).

Moreover, I can't seem to think of another way to quickly get an answer as this question is not meant to be too long since the whole thing should take about 25 minutes. I also get the sense that this construction must be explicit because of the very last part.

Now I should say that this is from an exam that was 23 years ago so there is a chance the syllabus has changed and that I am asking for something that is unreasonable.

If anybody sees a way to do (iv) without the spectral theorem I would greatly appreciate it if they could share.

Note: The only version of the Spectral Theorem that is allowed in the current version of the course is that if an ednomorphism is self adjoint then there is an orthonormal basis of eigenvectors of that endomorphism. I am sure that is not the full version of the theorem which seems to be used by most answers associated with this question on the website.

Remark: I would only like to be able to answer (iv) of the question.

Answer 1

Part (iii) tells us that $\alpha(E_\lambda^\perp) \subseteq E_\lambda^\perp$, but it is also true that $\alpha^*(E_\lambda^\perp) \subseteq E_\lambda^\perp$. One way to see this is that by (ii), a different definition of $E_\lambda$ is $\{x \in V : \alpha^*(x) = \overline \lambda x\}$, so we can apply (iii) from $\alpha^*$'s point of view instead of $\alpha$'s. Therefore $\alpha$ and $\alpha^*$, restricted to $E_\lambda^\perp$, are also endomorphisms of $E_\lambda^\perp$.

This lets us inductively solve the problem:

1. Pick an eigenvalue $\lambda$ of $\alpha$ (and $\alpha^*$);
2. Find an arbitrary orthonormal basis of $E_\lambda$.
3. If $E_\lambda = V$ we are done. (In particular, this must always happen if $V$ is $1$-dimensional, proving our base case.)
4. Otherwise, by strong induction on the dimension of $V$, we can find an orthonormal basis of eigenvectors of $\alpha$ restricted to $E_\lambda^\perp$; together with the orthonormal basis found in step 2, this gives us an orthonormal basis of all of $V$ consisting of eigenvectors of $\alpha$.