## Can always a family of symmetric real matrices depending smoothly on a real parameter be diagonalized by smooth similarity transformations?

This question is related to another question, but it is definitely not the same.

Is it always possible to diagonalize (at least locally around each point) a family of symmetric real matrices $A(t)$ which depend smoothly on a real parameter $t$? The diagonalization is supposed to be done by similarity transformations with a family of invertible matrices $S(t)$ depending smoothly on $t$.

An equivalent formulation is whether, given a smooth vector bundle $E\to \mathbb R$ over the one-dimensional differentiable manifold $\mathbb R$, and a smooth symmetric bilinear form $b\in E^*\otimes_M E^*$ on the vector bundle, we can always find a local frame of smooth sections of $E$ in which the symmetric bilinear form $b$ is diagonal.

Some results about such diagonalizations are known, for example from this article. There it is proven that we can smoothly choose the eigenvalues of a family of hermitian matrices depending smoothly on a real parameter, given that there are no points where the roots of the characteristic polynomial meet of infinite order. In particular, this works for the analytic case. But because the eigenvectors are computed with respect to the natural basis, it is implicitly assumed that they are orthogonal with respect to this basis.

My need is to find a smooth diagonalization by similarity transforms, which therefore are not necessarily orthogonal in respect to the natural basis, only invertible. Therefore, the restriction is weaker than that of having smooth eigenvectors.

Is it always possible to find such a diagonalization? If not, what are the conditions under which it can be done? Can you provide some references?

Thank you.

Update with the conclusions:

The counterexample of the type given by Michael Renardy and Denis Serre (who also explains it) answers my question (negatively). Initially I though of such examples as being valid only for the problem of eigenvectors, being of the type in the article to which I referred in my question, and I hoped that allowing the transformations to be more general than those orthogonal may avoid this problem.

But I understand now that the two problems are in fact equivalent. I think that the essence of the counterexample is to have the basis made of two rotating vectors on which the quadratic form corresponding to $b(t)$ has opposite signs. In this case, if by absurd we can diagonalize the matrices, even if $S(t)$ are not orthogonal, then in the new basis it is like solving the eigenvalue problem, and this possibility is contradicted by the counterexamples. For the positive definite case, Johannes Ebert pointed that the diagonalization is possible.

Thank you all for your help.

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 You say "the restriction is weaker than that of having smooth eigenvectors". I don't see this: if $S(t)^{-1} A(t) S(t) = D(t)$ where $D(t)$ is diagonal (say, just for $t$ near 0) then if $S(t)$ is smooth, then each column of $S(t)$ will be smooth. But the columns of $S(t)$ are just the eigenvectors of $A(t)$, with the associated eigenvalues being the entries of $D(t)$. – Matthew Daws Apr 4 2011 at 15:53 @Matthew Daws: You are right, I just realized this after thinking at the example from Kato's book, which appeared to me to be from the "less general" case. – Cristi Stoica Apr 4 2011 at 16:05 On the other hand, there is an "analytic SVD" (google it and you'll find several papers), which could suit your needs if you're thinking about a specific application. – Federico Poloni Apr 4 2011 at 19:39

A counterexample is given in Section II.5.3, p. 111 of T. Kato, Perturbation Theory for Linear Operators, 2nd ed.

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The question is whether you want to diagonalize an endomorphism or a bilinear form. Both problems are not equivalent (unfortunately sometimes confused in weaker linear algebra textbooks), diagonalizing bilinear forms is much easier. If the forms are positive definite, then the Gram-Schmidt process works with smooth families. If you have a trivial vector bundle $E \to X$ and a smooth positive definite bilinear form, then you can find a smooth orthonormal basis.

For self-adjoint operators, the problem is more subtle. Let $A(t)$, $t \in U$, $U$ a manifold, be a smooth family of self-adjoint endomorphisms of a Riemannian vector bundle $E \to U$. If the eigenvalues of $A(t)$ have multiplicity $1$ for each $t\in U$, then $E$ splits as a sum of line bundles each of which consists of eigenvectors of the family.

Proof idea: let $c_i$ be a small circle in the complex plane which encloses the eigenvalue $\lambda_i$ of $A(t_0)$ and no others. Then for $t$ close to $t_0$, $P_i (t):=\frac{1}{2 \pi i} \int_{c_i} (A(t)-z)^{-1} dz$ is an orthogonal rank $1$ projection commuting with $A(t)$, $P(t_0)$ projects onto the eigenspace for $\lambda_i$. For small $t$, $P(t)$ depends continuously on $t$. So the image of $P_i (t)$ for these $t$ is an eigenspace for $A(t)$.

Globally such a splitting exists as well, basically because the eigenvalues are real and therefore ordered. The line bundles may fail to be trivial, see Rolands example, where the $\pm 1$-eigenbundle is a MÃ¶bius bundle on $S^1$. For normal matrices, the eigenvalues can be complex and the splitting, though possible locally, might fail to exist globally. Counterexample: $B(t)=diag (exp(it),-exp(it))$, $C(t)=(cos(t/2),sin(t/2))\-sin(t/2),cos(t/2))$, $A(t)=C(t)^{-1} B(t) C(t)$, $A(0) = A(\pi)$ gives such a family on the circle.

If the multiplicities are greater than $1$, then eigenvalues are allowed to collide, which blows up the above argument. Counterexample: $U=(-1,1)$, $A(t)= diag (t,0)$ for $t \geq 0$, $A(t)=(0,t\t,0)$ for $t<0$ and smoothened around $t=0$.

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Set $c:=\cos\frac1t$ and $s:=\sin\frac1t$. Choose a function $\phi$ that is flat at $t=0$ ($\phi$ and all its derivatives vanish at $t=0$). Then set $$A(t):=\begin{pmatrix} \phi c^2 & -\phi sc \\ -\phi sc & \phi s^2 \end{pmatrix}.$$ This symmetric matrix is a $\mathcal C^\infty$-function of $t$. Yet its eigenvectors $(c, -s)^T$ and $(s,c)^T$ spin infinitely many times as $t\rightarrow0$.

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The answer is yes for real analytic or quasi-analytic parametrizations, see the following paper, also for an overview on different types of parametrizations: Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. pdf

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 Thank you for the reference. – Cristi Stoica Oct 2 at 18:00

Consider the symmetric matrices associated to orthogonal reflections of the Euclidean plane. Since a line comes back to itself after a half-turn, you get a family indexed by $[0,\pi]$ describing a closed path in the set of symmetric matrices. A corresponding eigenvector of eigenvalue $1$ and norm $1$, deformed continuously goes into its opposite. We get thus a closed path (of symmetric matrices) which does not give rise to closed paths of associated eigenvectors.

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