2
$\begingroup$

Let \begin{equation} A= \begin{pmatrix} 0 & a_{1,2} & a_{1,3} \\ a_{2,1} & 0 & a_{2,3} \\ a_{3,1} & a_{3,2} & 0 \end{pmatrix}, \end{equation}

\begin{equation} B= \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \end{equation} with $a_{i,j}>=0$ (no stochastic matrix) and $\pi\in\mathbb{R_{}}_{+}^3$. The system I want to solve is \begin{equation} \langle\pi,\pi B\rangle=\pi A \end{equation}, where $\langle.,.\rangle: \mathbb{R}^3\to\mathbb{R}^3$ is the elementwise multiplication of two vectors. Via programming, I found that with an arbitrary starting vector $\pi_0\in\mathbb{R}_{+}^3$, the following sequence converges to the desired solution $\pi$: \begin{equation} \pi_{i+1}:=\frac{\pi_{i}A}{\pi_{i}B}, \end{equation} where the fraction is again elementwise.

Can anyone help me prove existence and uniqueness of a solution (except the trivial one) to the system via this convergence? It looks a bit like the stationary distribution of irreducible Markov Chains, but it is different.

Also thankful for references to literature or other posts here. Thanks!

Remark: I already asked a quite similar question on Stackexchange Mathematics (https://math.stackexchange.com/questions/709568/how-to-solve-this-system-of-3-equations-with-3-variables) but did not get an answer. I thought MathOverflow would be more suited for the question. Please correct me, if I am wrong.

$\endgroup$
10
  • 1
    $\begingroup$ Are you interested in nonnegative solutions only? Otherwise, my guess is that the solution is not unique in general. $\endgroup$ Mar 22, 2014 at 12:27
  • $\begingroup$ there's a syntax error: $\langle \pi, \pi B\rangle$ is a scalar, whereas $\pi A$ is not... $\endgroup$
    – Suvrit
    Mar 22, 2014 at 12:40
  • $\begingroup$ @Federico: Thx, only nonnegative solutions. $\endgroup$ Mar 22, 2014 at 12:44
  • 1
    $\begingroup$ @Suvrit $\langle \pi, \pi B\rangle$ is not what you think, from the description it should be the Hadamard product. $\endgroup$ Mar 22, 2014 at 13:02
  • 1
    $\begingroup$ that's terrible notation then I'd say! $\endgroup$
    – Suvrit
    Mar 22, 2014 at 13:04

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.