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For a matrix $A$ I'm looking for a vector $\vec{\phi}$ such that: $$\frac {A\vec{\phi}}{tr(diag(\vec{A\phi}))}=\frac {\vec{\phi}}{tr(diag( \vec{\phi}))}$$

Or in other words: $$\frac{\sum_j a_{ij}\phi_j}{\sum_k\sum_j a_{kj}\phi_j}=\frac{\phi_i}{\sum_k \phi_k}$$

I made up the name for the problem calling it a relative fixed point, analogous to a fixed point problem $f(x)=x$.

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    $\begingroup$ these are the eigenvectors of $A$ $\endgroup$
    – Surb
    Apr 26 '18 at 17:31
  • $\begingroup$ the convergence you observe is simply the convergence of the power method $\endgroup$
    – Surb
    Apr 26 '18 at 17:45
  • $\begingroup$ convergence rate of this method is well studied. Take $A=\begin{pmatrix} 0&1\\1&0\end{pmatrix}$ and start at any point $v=(v_1,v_2)$ with $|v_1|\neq |v_2|$ see what happen. $\endgroup$
    – Surb
    Apr 26 '18 at 17:53
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Notice: OP has edited his question after my answer (the normalization factor...). Still a similar argument, shows that the solution to his edited equation are the eigenvectors $v$ of $A$ with eigenvalue $\lambda\neq 0$ and $\sum v_i \neq 0$.

Your equation characterizes the eigenvectors of $A$ with positive eigenvalue.

Indeed: Let $\|.\|$ be any norm (take $\|.\|=\|.\|_1$ if you want).
We say that $v$ is an eigenvector of $A$ if $v\neq 0$ and there exists $\lambda$ such that $$Av=\lambda v.$$ If $\lambda>0$, normalizing both sides of the equation, we have $$\frac{Av}{\|Av\|}=\frac{v}{\|v\|}$$ implying that $v$ solves your equation whenever $\lambda \neq 0$.

Now, if $w$ satisfes $$\frac{Aw}{\|Aw\|}=\frac{w}{\|w\|},$$ then $\|Aw\|,\|w\|>0$ and $$Aw=\frac{\|Aw\|}{\|w\|}w,$$ and so $w$ is an eigenvector with eigenvalue $\lambda=\frac{\|Aw\|}{\|w\|}\neq 0$.

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