# General solution of a vectorial ecuation

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?

• 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...) Jun 17, 2016 at 1:56
• Sorry, I didn't know that I had to ask in English. Here you are.
– Luks
Jun 17, 2016 at 2:12
• any eigenvector of $\boldsymbol{A}^t \boldsymbol{A}$ with norm $\sqrt{\frac{\phi}{1-q}}$ will be a solution for $z$. Jun 17, 2016 at 2:25
• 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!
– Luks
Jun 17, 2016 at 19:41

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.
• 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 $\| \boldsymbol{z} \|^2 = \frac{\phi}{1-q}$.
• . 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?