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I am reading about optimization and I am having difficulty in understanding the following:

If a matrix A is $n\times n$ Hermitian, then $\max_{x^{*}x=1} x^{*}Ax$ is solution equivalent to $\max_{x^{*}Ax=1} \frac{1}{x^{*}x}$.

Can someone help me understand how to arrive at this conclusion, or to show that they are indeed the same? Thanks.

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    That statement is only correct if the spectrum of $A$ has positive eigenvalues.2011-11-09

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Because $A$ is Hermitian, it has real spectrum. It is easy to see that $\max_{x^\ast x = 1} x^\ast A x = \lambda$. The maximum is attained for any unit norm $x$ vector from the eigenspace $V_\lambda$, corresponding to the eigenvalue $\lambda$.

Now, assuming $\lambda > 1$, let $y = \frac{1}{\sqrt{\lambda}} x$, then $y^\ast A y = 1$ and $\frac{1}{y^\ast y} = \lambda$. I hope you see the equivalence now, and what goes wrong if $A$ has no positive eigenvalue.