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Given a symmetric positive definite matrix $Q$, $$ \max_{\| x \| = 1} x^t Q x \tag{P} $$

Solving the problem I made the following:

  1. First I see that $\|x\| = 1$ is equivalent to $\|x\|^2=1$, so I replace it because it easier to use it.

  2. Then I construct the lagrange function $L(x,\lambda) = x^tQx +\lambda(\|x\|^2 -1)$

  3. I impose the gradient = 0:

$0=\nabla L(x,\lambda)=2Qx+2\lambda x = 2(Q+\lambda I)x$

So:

$Qx = -\lambda x$

So $-\lambda$ should be an eigenvalue of Q and then $-\lambda>0$

Even more if we are looking for the maximum taking

$-\lambda^* =$ max{μ_i eigenvalues}

And getting $x^*$ as any eigenvector of $-\lambda^*$ with $\|x^*\|^2=1$

We have that $(x^*)^tQx = -\lambda^*\|x\|^2 = -\lambda^*$ and that is maximum.

Is it the solution?

Rodrigo de Azevedo
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Pablo Diego Acosta
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  • It looks OK to me. – cs89 May 13 '23 at 19:27
  • Yes. You are correct. The result actually holds for any real symmetric matrix $Q$ (doesn't matter if it's positive definite): $\min_{|x| = 1} x^t Q x = \mu_\min$, $\max_{|x| = 1} x^t Q x = \mu_\max$. – nowhere May 13 '23 at 20:51
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    The problem can be solved without the Lagrange multiplier method. It suffices to use the following: if a matrix $A$ is positive definite then $x^tAx=0$ implies $Ax=0.$ Now to apply the above to the matrix $mI-Q,$ where $m=\max_{|x|=1}x^tQx.$ – Ryszard Szwarc May 13 '23 at 21:15

0 Answers0