## Eigenvalues of symmetric matrices II

[Updated 1/3/2011]
Some time ago I conjectured that for a smooth family of ${n\times n}$ symmetric real matrices ${A(x), x\in \Omega}$ (smooth here means all entries being smooth functions), there exists ${n}$ smooth families ${\lambda_1(x), \cdots, \lambda_n(x)}$ of eigenvalues of ${A(x)}$. As Raymond is asking me this question recently, I reconsider it again. Here I give an example that this is actually false. In my example, although locally we can always find ${n}$ (in fact, two) families of smooth eigenvalues, they cannot be continued in the whole domain to form even a single smooth family. (And also, my previous proof about ${n}$ continuous families of eigenvalues is also local in nature, I have corrected my post accordingly. )

The example is actually very simple: we consider a circle-family of matrices defined by

$\displaystyle A(\theta)= \begin{pmatrix} \sin \theta & 1-\cos \theta\\ 1-\cos \theta & -\sin \theta \end{pmatrix}, \theta\in [0, 2\pi).$

It is easy to see that ${A}$ can be regarded as a smooth family of symmetric matrices over the circle, and the characteristic polynomial ${\det (A-\lambda I)}$ is

$\displaystyle \lambda^2-4\sin^2(\frac{\theta}{2}).$

So we actually have two continuous families of eigenvalues: ${\pm 2 \sin(\frac{\theta}{2})}$. Also, as ${\pm 2\sin(\frac{\theta}{2})}$ are distinct if ${\theta\neq 0}$, if there are two families of smooth eigenvalues, they must be ${\pm 2 \sin(\frac{\theta}{2})}$ by continuity. However, it can be easily seen that ${2\sin(\frac{\theta}{2})}$ is not a smooth function (at ${\theta=0}$) when regarded as a function over the circle, thus there are no such families. Actually the graph of the two families look something like the (boundary of the) “Mobius band” over the circle: they cross each other at ${\theta=0}$, and actually, near ${\theta=2\pi^-}$, the family ${2\sin(\frac{\theta}{2})}$ can be extended smoothly beyond ${2\pi}$, but by taking the value ${-2 \sin(\frac{\theta}{2})}$ instead (i.e. the “second” family). Thus, we can still locally find two such smooth families of eigenvalues.

It is easy to see that the above circle family of matrices can be extended to be a family of matrices over a (annular, say) domain in ${\mathbb{R}^2}$, if you want. My new conjecture is: given a smooth (whatever reasonable sense) family of ${n\times n}$ real symmetric matrices ${A}$, we can locally find ${n}$ families of smooth eigenvalues of ${A}$.

***

Some words about the construction of the above matrix.

Although the matrix above looks very simple algebraically, I do not arrive at it very directly. I am not used to think “algebraically”, so I construct the example geometrically as follows.

The picture is simple: construct circle-family of linear maps whose eigenvalues look like the “Mobius band” as described above.

My method is: for some fixed (global) orthonormal frames ${e_1, e_2}$ over the circle, take ${v_1(\theta)=\cos (\frac{\theta}{4})e_1+\sin(\frac{\theta}{4})e_2}$ and ${v_2(\theta)=-\sin(\frac{\theta}{4})e_1+\cos(\frac{\theta}{4})e_2}$ for ${\theta\in[0, 2\pi)}$. Note that ${v_1(\theta)}$ will “become” ${v_2(0)}$ and ${v_2(\theta)}$ will “become” ${-v_1(0)}$ after traveling one loop (i.e. the limit at ${2\pi}$). I then take the map ${A: v_1\mapsto \sin (\frac{\theta}{2})v_1, v_2\mapsto -\sin (\frac{\theta}{2})v_2}$ for ${\theta\in [0, 2\pi)}$. It is easy to see that the two eigenvalues (i.e. ${\pm \sin\frac{\theta}{2}}$) crosses each other at ${\theta=0}$ and also ${A}$ is continuous also at ${\theta=0}$. It is actually not hard to see (geometrically) that ${A}$ is smooth at ${\theta=0}$. With some effort, we calculate the matrix representation of the map with respect to ${\{e_i\}}$ is the above matrix.

***

There is also a related question raised by Raymond.

Question: Suppose ${A(x)}$ is a smooth (in any reasonable sense) family of real symmetric matrices, and suppose that we now have a smooth family of eigenvalue ${\lambda(x)}$, does it follow that we can find a smooth family of eigenvectors ${v(x)}$ w.r.t. ${\lambda(x)}$? (Recall that an eigenvector can’t be zero. )

Answer: No. Similar to the above question, the global existence of such family depends heavily on the topology of the domain (rather than linear algebra).

Take a sphere (${\mathbb{S}^2}$) family of matrices

$\displaystyle A(p)= \begin{pmatrix} p_1 \vec p & p_2 \vec p & p_3 \vec p \end{pmatrix}$

where ${\vec p=(p_1, p_2, p_3)\in \mathbb{S}^2}$ (regarded as a column vector). It is easy to see that ${A}$ is symmetric and geometrically ${A}$ is the projection into the normal direction: i.e. ${A(p)\vec x=\langle x, p\rangle p,}$ noting that ${p}$ can be regarded as the normal vector of the sphere at ${p}$. Thus ${A}$ has ${1}$ as an eigenvalue (with the normal vector ${p}$ as eigenvector) and an eigenvalue ${0}$ with multiplicity ${2}$ (corresponds to the two dimensional tangent space of ${\mathbb{S}^2}$, as they have no normal component). Take ${\lambda(p)\equiv 0}$, finding ${v(p)}$ amounts to finding a global non-vanishing vector field on the sphere. But according to the Poincare-Hopf theorem (or its friendlier child, Hairy ball theorem), which states that the index of any tangent vector field on the sphere must be 2, it follows that any tangent vector field on ${\mathbb{S}^2}$ must have a zero somewhere, therefore no such ${v(p)}$ exists.

[Updated 1/3/2011]
According to Terence Tao (see here), the eigenvalues of a symmetric matrix should be smooth, locally, by Rellich’s theory. However, I don’t have the knowledge of that theory. Anyone familiar with it?