TITLE: Proof of Spectral Theorem
QUESTION [8 upvotes]: There are a number of results called Spectral Theorems. This question deals with the Linear Algebra result on normal operators, which has the self-adjoint case as a particular case.
In class, we saw the spectral theorem for self-adjoint operators, and the teacher attempted a sketch of a proof which nobody in the class understood. Writing my thesis, I hit into the normal operator case, so I was looking for a complete proof of it.
How is it proven?

REPLY [2 votes]: The spectral theorem says that every normal operator$~\phi$ on a finite dimensional complex inner product space$~V$ is diagonalisable, and that its eigenspaces are mutually orthogonal. As a consequence an orthonormal basis of$~V$ consisting eigenvectors for$~\phi$ can be chosen.
Here is a simple proof. I will start with the special case where $\phi$ is an Hermitian (also called self-adjoint) operator, which is quite easy; more so indeed than the case of a self-adjoint-operator on a finite dimensional real inner product space (also called Euclidean space) that I discussed here. Then I will use this result to generalise it to the case of normal operators.
A basic fact about adjoints is that for any linear operator $\phi$ on$~V$, whenever a subspace $W$ is stable under$~\phi$, its orthogonal complement $W^\perp$ is stable under its adjoint$~\phi^*$. For if $v\in W^\perp$ and $w\in W$, then $\langle w\mid \phi^*(v)\rangle=\langle \phi(w)\mid v\rangle=0$ since $\phi(w)\in W$, so that $\phi^*(v)\in W^\perp$. Then for a Hermitian operator $\phi$ (so with $\phi^*=\phi$), the orthogonal complement of any $\phi$-stable subspace is again $\phi$-stable.
Now we prove the spectral theorem for Hermitian $\phi$ by induction on $\dim V$. When $\dim V=0$ the unique operator $\phi$ on $V$ is diagonalisable with empty set of eigenvalues, and the result is trivial. Now assuming $\dim V>0$, there is at least one eigenvalue (since the characteristic polynomial of $\phi$ has a root by the fundamental theorem of algebra) so we can choose an eigenvector $v_1$ of$~\phi$. The subspace $W=\langle v_1\rangle$ it spans is $\phi$-stable by the definition of an eigenvector, and so $W^\perp$ is $\phi$-stable as well. We can then restrict $\phi$ to a linear operator on $W^\perp$, which is clearly self-adjoint, so our induction hypothesis gives us an orthonormal basis of $W^\perp$ consisting of eigenvectors for that restriction; call them $(v_2,\ldots,v_n)$. Viewed as elements of $V$, the vectors $v_2,\ldots,v_n$ are eigenvectors of$~\phi$, and clearly the family $(v_1,\ldots,v_n)$ is orthonormal. It is an orthonormal basis of eigenvectors of$~\phi$, and we are done.

So now I will deduce from this the more general case where $\phi$ is a normal operator. First, note that the from Hermitian case easily follows the anti-Hermitian case, i.e., the one where $\phi^*=-\phi$. One way is to observe that for anti-Hermitian $~\phi$ one still has that if a subspace $W$ is $\phi$-stable then so is its orthogonal complement $W^\perp$ (just put a minus sign in the above argument), so the proof of the spectral theorem for Hermitian$~\phi$ can be reused word-for-word. Another way is to observe that $\phi$ is Hermitian if and only if $\def\ii{\mathbf i}\ii\phi$ is anti-Hermitian, so if one is diagonalisable so is the other, with the same eigenspaces but with the eigenvalues of $\ii\phi$ multiplied by$~\ii$ with respect to those of$~\phi$. (One easily shows the eigenvalues are real in the Hermitian case and imaginary in the anti-Hermitian case, but that is of no importance here.)
Next observe that any linear operator $\phi$ can be written as the sum of a Hermitian operator $\frac12(\phi+\phi^*)$ and an anti-Hermitian operator $\frac12(\phi-\phi^*)$, which are called its Hermitian and anti-Hermitian parts. Moreover $\phi$ and $\phi^*$ commute (i.e., $\phi$ is a normal operator) if and only if the Hermitian and  anti-Hermitian parts of $\phi$ commute.
So for a normal operator$~\phi$ we know that its Hermitian and anti-Hermitian parts are both diagonalisable with mutually orthogonal eigenspaces, and the two parts commute. Now apply the well known fact that two commuting diagonalisable operators can be simultaneously diagonalised; any basis of common eigenvectors is a basis of diagonalisation for$~\phi$. Finally, if one has two eigenvectors of$~\phi$ for distinct eigenvalues, then they are also eigenvectors for the Hermitian and anti-Hermitian parts of$~\phi$, and for at least one of the two they are so for distinct eigenvalues; by virtue of that fact the vectors are orthogonal. This shows that eigenspaces of$~\phi$ are mutually orthogonal.

In fact, it remains true for a normal operator$~\phi$ that the orthogonal complement of a $\phi$-stable subspace is always $\phi$-stable, so one could have reused the proof of the Hermitian case once again. However to prove this fact would seem to be more work than the separate argument I gave.