# How to find all stationary points of $\alpha\|v\|^2-\|x^Tv\|^2+\|g^Tv\|^2$

Let $v,x,g$ be three vectors and $\alpha$ be a constant. The problem is

$$\min\limits_v \{\alpha\|v\|^2-\|x^Tv\|^2+\|g^Tv\|^2\}$$

where $\|v\|^2=\sum\limits_{i=1}^{|v|}v_i^2$ and $|v|$ is the cardinality of $v$.

$$\alpha v-x^Tvx+g^Tvg.$$

How should I get all vector $v^*$ such that $\alpha v^*-x^Tv^*x+g^Tv^*g=0$?

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Your equation is equivalent to : $$(\alpha I - xx^T+gg^T)v=0$$ So $v$ should be in the kernel of matrix ($\alpha I -xx^T+gg^T)$ (which might not exist). So it must be a linear combination of vectors $x$ and $g$. Define $v= ax+bg$. Then plugging this parametrization in the equation you will get equation for $a,b$ which are defined up to multiplier. But ...taking the gradient it is not a correct way to solve the problem. Without the constraint on $v$ it might be not well defined. It depends on $\alpha,x,g$. – Alexander Vigodner Jul 2 '14 at 20:38
I meant the problem is not well defined. Gradient of course can be calculated anyway. Assume that $v^*$ is an optimal solution. Then if $\min$ is positive it can be decreased to zero setting $v=0$. So $v^*$ is not optimal. If $\min$ is negative then it can be set to $-\infty$ increasing the vector. So you must set a constraint on $v$. – Alexander Vigodner Jul 2 '14 at 21:12

Obviously, there is no closed form solution. But what you can do is to solve the system $v=1/\alpha*(x^Tvx-g^Tvx)$. One way to solve is to iterative update $v$ from a initial point $v^0$. It is guarantee to converge but the result is only one local solution. If you want to obtain all $v$, it seems impossible.