I want to know on how do I prove this following statement of the Rayleigh Quotient. If A is symmetric, the optimization values (I) and (II) below have the same optimal value. If A has at least one positive eigenvalue, then the problems (I) to (III) have same optimal value

(I) $\max_{x^{\top}x=1} x^{\top}Ax$

(II) $\max_{x\neq0} \frac{x^{\top}Ax}{x^{\top}x}$

(III) $\max_{x^{\top}Ax=1} \frac{1}{x^{\top}x}$

Hint: If A is negative semi-definite, there is no x verifying the constraint of the problem (III)

The only thing I know is symmetric matrix is a square matrix that is equal to the transpose and the Rayleigh Quotient which is (II) not including the max part and Semi-definite is a symmetric matrix denoted by A $\ge$ 0 if $x^{\top}Ax$ $\ge$0 for every x $\in$ R

Is it possible you can give me any hints on how to start this question? thank you.

  • 1
    $\begingroup$ Didi you mean [$\frac{1}{x^{\top}x}$] instead of [$\frac{1}{x^{\top}}$] for (III) ? $\endgroup$ – Duchamp Gérard H. E. Apr 21 '16 at 10:46

Some hints : (I) and (II) have the same value because you can rescale any $x\not=0$ to a unitary vector $u$ by $$ u=\frac{x}{\sqrt{x^\top x}} $$ As the sphere of equation $x^\top x=1$ is compact, the max of (I) is reached at, say $x_0$ and is strictly positive (because your matrix has at least a strictly positive eigenvalue), then, again, you can rescale your vector $x_0$ and get (III). Does it help ?

  • $\begingroup$ Yes it does help thank you sorry for the late reply $\endgroup$ – TriariiD Jun 3 '16 at 6:26

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