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This is the first time that I ask a question here. When I was looking for the maximum of a multivariable vector function, I encountered the following problem: I cannot find the general solution of the following vectorial equation:

$$ \boldsymbol{A}^t \boldsymbol{A} \boldsymbol{z} - \frac{1-q}{\phi} \boldsymbol{z}^t \boldsymbol{A}^t \boldsymbol{A} \boldsymbol{z} . \boldsymbol{z} = \boldsymbol{0} , $$

where the matrix $\boldsymbol{A} \in \mathbb{R}^{p \times (p+1)}$, the unknown vector $ \boldsymbol{z} = (x, \boldsymbol{y}^t)^t \in \mathbb{R}^{p+1}$ given that $x \in \mathbb{R}$ and $\boldsymbol{y} \in \mathbb{R}^{p})$, $q \in (0 , 1)$ and $\phi >0$.

I could solve it for the case $ p =1$. In this case $\boldsymbol{A} = (a_1, a_2)$, $\boldsymbol{z} = (x, y)^t$ and the solutions are the point $ \sqrt{\frac{\phi}{(1-q) (a_1^2 + a_2^2)}} . (a_1, a_2)^t$, its opposite and all the points of the line $ y = - \frac{a_1}{a_2} x$.

My problem is that no matter how much I try I cannot generalize it for $p>1$. Any advice?

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    $\begingroup$ Downvoter: seriously? I can't be certain, but it looks better than quite a few English questions! Surely there are Spanish-speaking people around here, and one of them might be kind enough to translate (it's currently better than garbled English like a poor translation attempt, or Google translate...) $\endgroup$
    – pjs36
    Jun 17, 2016 at 1:56
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    $\begingroup$ Sorry, I didn't know that I had to ask in English. Here you are. $\endgroup$
    – Luks
    Jun 17, 2016 at 2:12
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    $\begingroup$ any eigenvector of $\boldsymbol{A}^t \boldsymbol{A}$ with norm $\sqrt{\frac{\phi}{1-q}}$ will be a solution for $z$. $\endgroup$ Jun 17, 2016 at 2:25
  • $\begingroup$ You are right. And adding the solutions for the zero eigenvalue (that is to say, all the $\boldsymbol{z}$ such that $ \boldsymbol{A^t A} \boldsymbol{z} = \boldsymbol{0} $), I think I have all the solutions to my problem. Thankyou! $\endgroup$
    – Luks
    Jun 17, 2016 at 19:41

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All you are asking for is the eigenvectors of $B = A^T A,$ which is symmetric positive semidefinite but has rank at most $p,$ so there is at least one null vector. There is still an (orthogonal) basis of eigenvectors.

Once you have $Bz = \lambda z,$ you get $$ Bz = \frac{\lambda}{z^TBz} (z^T B z) z $$

That is when $\lambda \neq 0.$ When $\lambda = 0,$ you get $BZ = 0$ and that's it.

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  • $\begingroup$ Ok. So, for the eigenvalue $\lambda =0$ I have the solutions of the lineal sistem $ \boldsymbol{A}^t \boldsymbol{A} \boldsymbol{z} = \boldsymbol{0}$ as solution to my problem. And for the eigenvalues $\lambda \neq 0$ I have the eigenvectors $\boldsymbol{z}$ asociated to $\lambda$ such that $ \| \boldsymbol{z} \|^2 = \frac{\phi}{1-q} $. $\endgroup$
    – Luks
    Jun 17, 2016 at 19:52
  • $\begingroup$ . I think those are all the solutions to the original equation. My target is to find the maximum value of the function $f(\boldsymbol{z}) = \exp \left(-\frac{1-q}{\phi} \| \boldsymbol{z} \|^2 \right) \boldsymbol{z}^t \boldsymbol{A}^t \boldsymbol{A} \boldsymbol{z}$. Thank to your help, I can see now that the maximum value that f can get wll be $ \frac{\phi \lambda_0}{e (1-q)} $, where $\lambda_0$ is the greatest eigenvalue of $\boldsymbol{A}^t \boldsymbol{A}$. Am I right? $\endgroup$
    – Luks
    Jun 17, 2016 at 20:10

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