# Relative Fixed point vector of a matrix

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 $f(x)=x$.

• these are the eigenvectors of $A$
– Surb
Apr 26 '18 at 17:31
• the convergence you observe is simply the convergence of the power method
– Surb
Apr 26 '18 at 17:45
• 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.
– Surb
Apr 26 '18 at 17:53

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 $$\lambda=\frac{\|Aw\|}{\|w\|}\neq 0$$.