Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

2 Answers 2

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).

share|improve this answer

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 $\dim \Delta = (n^2-n-2)+(n-1) = n^2-3$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.