# 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.

• Note that in general, we should not expect the limit of diagonalizable matrices to be diagonalizable. In particular, note that $$\pmatrix{1 + 1/n & 1\\0 & 1}$$ is diagonalizable for all $n \in \Bbb N$, but its limit is not diagonalizable. – Omnomnomnom Sep 24 '17 at 20:53

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.

• Thanks for your answer! Density is sufficient here. – Vim Sep 25 '17 at 11:19
• I think the space of hermitians should be $n(n-1)/2$-dimensional instead. – Vim Sep 25 '17 at 12:46
• @Vim: You can see that your dimension count is false already for $n = 2$. A Hermitian matrix depends on $n$ real parameters (diagonal entries $a_{ii}$) and $1 + 2 + \dots + (n-1) = \frac{n(n-1)}{2}$ complex parameters $a_{ij}$ for $i > j$ which results in a total of $n + 2 \frac{n(n-1)}{2} = n^2$ real parameters. – levap Sep 25 '17 at 13:23
• alrite I was thinking over $\Bbb C$... but still a silly mistake. – Vim Sep 25 '17 at 13:25
• Note that it is not a vector space over $\mathbb{C}$. – levap Sep 25 '17 at 13:25

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 $U$.

• Even slicker: if $D_{n,B}$ is the set of diagonal matrices with entries bounded by $B$, then $D_{n,B}\times U_n$ is compact, and thus the set of all $U^*DU$ is closed since it's the image of a comapct set under a continuous function. – Hurkyl Sep 24 '17 at 22:15
• Thanks for your answer. To clarify a detail: when you are talking about $\Bbb C^{d\times d}$, what's the topology you're using here? I would assume it is the topology induced by Frobenius norm (i.e. "Entrywise" topology). – Vim Sep 25 '17 at 0:01
• @Vim That works. As it turns out, every normed topology on $\Bbb C^{d \times d}$ is equivalent to the "entrywise" topology. – Omnomnomnom Sep 25 '17 at 0:32