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Let $A \in \mathbb{R}^{n \times n}$ be a positive definite matrix and let $a > 0$ be some constant.

I am interested in the following quantity $$ \hat x = {\rm arg} \max_{x \in R^n}\ x' \left( \frac{1}{a + \|x\|^2_2} A \right) x $$ Is it possible to relate $\|\hat x\|_2^2$ to the eigenvalue and $\hat x$ to the eigenvector of $A$.

Furthermore, suppose that $E \in \mathbb{R}^{p \times p}$ is such that $A + E$ is still positive definite and $\|E\|$ is small in some norm (say, spectral). Consider $$ \hat y = {\rm arg} \max_{y \in R^n}\ y' \left( \frac{1}{a + \|y\|^2_2} (A+E) \right) y $$ Are there known results that can be used to quantify $| \|\hat x\|_2^2 - \|\hat y\|_2^2 |$? I suppose the norms should be close if the perturbation $E$ is small, but don't know how to quantify this notion.

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The minimum is always attained for x = 0. Are you missing something or is it me? – Michael Ulm Nov 9 '10 at 11:42
Indeed, the minimum is always attained. I meant to ask about the maximum. Thanks for pointing this out. – mkolar Nov 9 '10 at 11:59
up vote 3 down vote accepted

Let $\lambda$ be the biggest eigenvalue of A and $x_{\lambda}$ a corresponding eigenvector. Since $x' A x \leq \lambda \left\| x \right\|^2$, we obtain

$$ x' \left( \frac{1}{a + \left\| x \right\|_2^2} A \right) x \leq \frac{\lambda \left\| x \right\|^2}{a + \left\| x \right\|^2} < \lambda $$

But for $x = t x_{\lambda}$ and $t \to \infty$, the expression goes to $\lambda$.

This means the maximum is never attained.

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