Trying to formalise intuition into a proof that symmetric (hermitian) matrices are diagonalisable The other day I stumbled upon LittleO's answer in this question. To make reading it easier I did a straight copy-and-paste here:

Here's some intuition (but not a rigorous proof).
If $A$ is hermitian (with entries in $\mathbb C$), you can easily show that the eigenvalues of $A$ are real and that eigenvectors corresponding to distinct eigenvalues are orthogonal.
Typically, all the eigenvalues of $A$ are distinct.  (It is in some sense a huge coincidence if two eigenvalues turn out to be equal.) So, typically $A$ has an orthonormal basis of eigenvectors.
Even if $A$ has some repeated eigenvalues, perturbing $A$ slightly will probably cause the eigenvalues to become distinct, in which case there is an orthonormal basis of eigenvectors.  By thinking of $A$ as a limit of slight perturbations of $A$, each of which has an ON basis of eigenvectors, it seems plausible that $A$ also has an ON basis of eigenvectors.

While this idea struck me as very ingenious and intuitive, I'm really having difficulty formalising it into a real proof. The major obstacles are:
1). How to perturb an hermitian $A$ while keeping its eigenvalues distinct? In particular, how to find an hermitian sequence $A_n$ that have distinct eigenvalues and converge to $A$ in some norm? 
2). Based on 1), how to continuously associate the eigenvector family $V_n:=[v_{n,j},j=1,\cdots,d]$ (corresponding to $A_n$, assuming $A_n$ are $d$ by $d$) to $A_n$? Having solved this I would have $V:=[v_j,j=1,\cdots,d]$ (corresponding to $A$) are the limits of $\{v_{n,j},j=1,\cdots,d\}$ respectively and hence $V^HV = \lim_{n\to\infty} V_n^HV_n = I$ and we are done.
 A: In the below, I use $A^*$ to denote the conjugate transpose of $A$.
In truth, I'm not sure how to answer part 1.  In particular, I'm finding it difficult to prove the existence of such a perturbation in such a fashion that doesn't amount to what is essentially a full proof of the spectral theorem.
It is notable, however, that if $\lambda,v$ is an eigenpair of $A$ and $\|v\| = 1$, then the matrix $B = A + \epsilon vv^*$ (where $\epsilon > 0$ can be made arbitrarily small) has "all the same eigenvectors" and will satisfy


*

*$Bu = Au$ for $u \perp v$

*$Av = (\lambda + \epsilon)v$


Moreover, it is straightforward to show that $A(vv^*) = (vv^*)A$.

Assuming that part 1 is accounted for, there is a neat argument for part 2.  Presumably, we have a sequence of Hermitian matrices $A_n \to A$ such that each $A_n$ has distinct eigenvalues, and we have an associated sequence of unitary matrices $U_n$ such that $U_n^*A_nU_n$ is a diagonal matrix.  
Notably, the set of unitary matrices is compact in $\Bbb C^{d \times d}$.  
So, there exists a convergent subsequence $U_{n_k}$.  Let $U$ denote the limit of this subsequence.  Notably, we have $A_{n_k} \to A$, since the subsequence of a convergent sequence is convergent with the same limit.  From there, we note that matrix-multiplication and the $A \mapsto A^*$ operator are continuous to conclude that
$$
U^*AU = 
\left( \lim_{k \to \infty} U_{n_k}\right)^*
\left( \lim_{k \to \infty} A_{n_k}\right)
\left( \lim_{k \to \infty} U_{n_k}\right)
\\ 
=
\left( \lim_{k \to \infty} U_{n_k}^*\right)
\left( \lim_{k \to \infty} A_{n_k}\right)
\left( \lim_{k \to \infty} U_{n_k}\right)
\\ 
= \lim_{k \to \infty} (U_{n_k}^*A_{n_k}U_{n_k})
$$
So, $U^*AU$ is the limit of a sequence of diagonal matrices in $\Bbb C^{d\times d}$.  Since the set of diagonal matrices is closed in $\Bbb C^{d \times d}$, we may conclude that this limit is also diagonal.  In other words, the Hermitian matrix $A$ is diagonalized by the unitary matrix $U$.
A: Let me answer part $(1)$ as part $(2)$ was answered by Omnomnomnom. Consider the space $W$ of Hermitian $n \times n$ matrices as a real vector space (which is isomorphic to $\mathbb{R}^{n^2}$) and let $D \colon W \rightarrow \mathbb{R}$ be the map which maps a Hermitian matrix $A$ to the discriminant of the characteristic polynomial of $A$. The map $D$ is polynomial in the entries of the matrix $A$ and $D(A) \neq 0$ if and only if the roots of the characteristic polynomial of $A$ are distinct. For example, if $n = 2$, a Hermitian matrix is described as
$$ A = \begin{pmatrix} x & a + ib \\ a - ib & y \end{pmatrix} $$
for $a,b,x,y \in \mathbb{R}$ and
$$ D(A) = (x + y)^2 - 4(xy - |a + ib|^2) = (x - y)^2 + (2a)^2 + (2b)^2. $$
Since $D$ is a non-zero polynomial (it doesn't vanish on the diagonal matrices with distinct eigenvalues), the set $\{ A \in W \, | \, D(A) \neq 0 \}$ is dense in $W$ so any Hermitian matrix in $D^{-1}(0)$ can approximated by Hermitian matrices with distinct eigenvalues.
In fact, one can show that the set $D^{-1}(0)$ of Hermitian matrices with a repeated eigenvalue is of codimension three in $W$ (see "avoidance of crossings" in Lax's Linear Algebra) but I don't know how to do it without the spectral theorem.
