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Suppose that we have a family of linear functions $L(\alpha) : \mathbb{R}^n \rightarrow \mathbb{R}^n$, where $\alpha$ is a positive real number.

For each $\alpha$, it is given that $L(\alpha)$ is a symmetric matrix, and so it has a basis of eigenvectors, and $n$ eigenvalues (not necessarily distinct). We define a function $f(\alpha)$ to be the smallest eigenvalue of $L(\alpha)$. For example, if $L(\alpha)$ is the matrix

$ \left( \begin{array}{cc} \alpha & 0 \\ 0 & 2\alpha \end{array} \right) $

then $f(\alpha) = \alpha$ because $\alpha > 0$. It is also given that the elements of $L(\alpha)$ are smooth functions of $\alpha$ for each $\alpha$. Therefore, as the eigenvalues are the roots of a polynomial whose coefficients are also smooth functions of $\alpha$, $f(\alpha)$ is a smooth function of $\alpha$ for all but a few values of $\alpha$ (this follows from the implicit function theorem).

The problem I am having is to try to find the global minimum of $f(\alpha)$. Usually what one would do is to take its derivative with respect to alpha, and set that equal to 0, but that is actually not possible in the present case because there is no way to get an explicit formula for $f(\alpha)$.

A possible idea that I had was to get some kind of sequence which converges to the smallest eigenvalue of $L(\alpha)$, and hope that this sequence converges uniformly in $\alpha$, but I have not yet found such a thing.

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You can't use Rayleigh-Ritz somehow? The smallest eigenvalue is $\min_{\| x \| = 1} x^T A x$ for real symmetric $A$. – Calle Feb 16 '11 at 2:54
@Calle: I've never heard of that, could you give me a reference to it? I tried looking it up on wikipedia but I couldn't find that specific theorem. – Eric Haengel Feb 16 '11 at 5:25
It can be seen as a special case of the min-max theorem, – Calle Feb 16 '11 at 14:12
up vote 0 down vote accepted

Always a fan of the cheap solution, suppose I take $n=1$, then $f(\alpha)$ has just one eigenvalue. And $\frac{df(\alpha)}{d\alpha}$ can be anything.

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Thank you for your hard work! lol – Eric Haengel Feb 16 '11 at 5:24
But that applies to other dimensions as well. Just extend it with 1's down the main diagonal and 0's elsewhere. So I think you need some more conditions. – Ross Millikan Feb 16 '11 at 5:32

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