# Eigenvalues of a Hermition Matrix do not cross

Wikipedia's article on avoided crossing asserts that "The eigenvalues of a Hermitian matrix depending on N continuous real parameters cannot cross except at a manifold of N-2 dimensions."

If it's true, does anyone have an elegant proof of this statement? Does anyone have a good interpretation or a good intuition for why this would be true?

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That cannot be right in this general form. See $$\begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix}$$ for $x, y \in \mathbb{R}$. Eigenvalues cross at $x=y$ of co-dimension one, no? – WimC Apr 29 '13 at 7:53

It is treated in detail in Lax's book on linear algebra. It is a statement about what happens generically (and it is easy to produce counterexamples to the general statement).

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As WimC says, the eigenvalues of $\left( \begin{smallmatrix} x & 0 \\ 0 & y \end{smallmatrix} \right)$ collide in codimension one. So this is not always a true statement.

On the other hand, let $H$ be the space of $n \times n$ Hermitian matrices, which is a real vector space of dimension $n^2$. Let $\Delta \subset H$ be the locus where two of the eigenvalues collide. Then $\dim \Delta = n^2 -3$. If we have a family of Hermitian matrices parameterized by some parameter set $B$, then we get a map $\phi: B \to H$ and we are interested in the dimension of $\phi^{-1}(\Delta)$. Generically, we should expect the image of $\phi$ to be transverse to $\Delta$, so we should expect $\phi^{-1}(\Delta)$ to be codimension $3$. I can't see where Wikipedia is getting the codimension $2$ number.

Computation of the dimension of $\Delta$: The unitary group $U_n$ acts on $H$ and $\Delta$ is a union of $U_n$ orbits. The subset of $\Delta$ where precisely two of the eigenvalues collide is dense in $\Delta$. So let's compute the dimension of the space where precisely two eigenvalues collide.

The space of matrices with eigenvalues $\lambda_1$, $\lambda_2$, ..., $\lambda_{n-2}$, $\mu$, $\mu$ is a $U_n$ orbit. The stabilizer of $\mathrm{diag}(\lambda_1, \lambda_2, \ldots, \lambda_{n-2}, \mu, \mu)$ is $U(1)^{n-2} \times U(2)$, of dimension $(n-2) + 4=n+2$. So the size of the orbit is $n^2 - n -2$. There are $n-1$ ways to choose the parameters $\lambda_1$, $\lambda_2$, ..., $\lambda_{n-2}$, $\mu$. So $\dim \Delta = (n^2-n-2)+(n-1) = n^2-3$.

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The dimension of $U(2)$ is $3$, not $4$. – WimC Apr 29 '13 at 13:45
No, $SU(2)$ is $3$ dimensional, $U(2)$ is $4$ dimensional. When $n=2$, the space of hermitian matrices with equal eigenvalues is $1$ dimensional (scalar multiples of the identity), so codimension $3$. – David Speyer Apr 29 '13 at 13:48
You are absolutely right of course. Sorry for the noise... – WimC Apr 29 '13 at 13:50
Generic $U(n)$ orbits are only $(n^2-1)$-dimensional since the center acts trivially (the trace of $H$ is preserved). So the dimension of $\Delta$ is $n^2-2$ after all. Right? – WimC Apr 29 '13 at 14:16
The center is included in the $(n-2)+4$ dimensional stabilizer I described. Again, think about the $n=2$ case. (Also, generic orbit have dimension $n^2-n$; all the coefficients of the characteristic polynomial are preserved.) – David Speyer Apr 29 '13 at 15:18