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Given a diagonal matrix $\Lambda$ (with at most one entry equal to zero) and an arbitrary matrix $M$, is it possible to determine $S$ from the matrix equation $S = M + \Lambda S \Lambda$?

I do not have any experience in solving equations like this, so any suggestions on how to start would be most welcome. Recommendations for (introductory) literature on the topic would also be much appreciated.

Ailurus
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1 Answers1

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This is very simple because $\Lambda$ is diagonal.

Write out $$ S = M + \Lambda S \Lambda $$ in coordinates $$ S_{ij} = M_{ij} + \sum_{kl} \Lambda_{ik} S_{kl} \Lambda_{lj} $$ Because $\Lambda$ is diagonal than this simplifies to $$ S_{ij} = M_{ij} + \Lambda_{ii} S_{ij} \Lambda_{jj} $$ Form this you get $$ S_{ij} = \frac{M_{ij}}{ 1- \Lambda_{ii}\Lambda_{jj}} $$ And you can see the condition under which your equation has a solution $$\Lambda_{ii}\Lambda_{jj} \neq 1 \qquad \text{or} \qquad M_{ij} = 0$$

tom
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  • Thanks! Hmm, $\Lambda_{11} = 1$, all other $\Lambda_{ii} < 1$. However, the value of $S_{11}$ (in fact the entire first column $S_{i1}$) seems to be irrelevant in my case, since $S$ is eventually post-multiplied by a column vector whose first entry is $0$ (I didn't mention this because it didn't seem relevant). – Ailurus Apr 17 '14 at 19:05
  • And what is $M_{11}$? If it is zero, than you are fine. The condition should be a little different, I edited the answer. – tom Apr 17 '14 at 19:41
  • Yes, $M_{11}$ happens to be zero (in fact the entire first row of $M$ is zero, but the other bits of $M$ are arbitrary). This means that $S_{11} = \frac{0}{0}$, but I don't see why this would be defined as zero in this case? – Ailurus Apr 17 '14 at 20:37
  • Well the equation for $S_{ij}$ is $(1- \Lambda_{ii}\Lambda_{jj})S_{ij} = M_{ij}$. Only when $1- \Lambda_{ii}\Lambda_{jj}$ is nonzero, than you can divide by it and obtain $S_{ij} = \frac{M_{ij}}{ 1- \Lambda_{ii}\Lambda_{jj}}$. If both $M_{ij}$ and $1- \Lambda_{ii}\Lambda_{jj}$ is zero than $S_{ij}$ can be anything. – tom Apr 17 '14 at 23:05