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I have a arbitrary $N \times N$ matrix $S$. I have an function of this matrix given by:

$A = F(S) = 2S + P^{-1}S + 2SP + PSP$

where $P$ is the a cyclic permutation matrix which when acting on the left of a matrix sends row $i$ to row $i-1$ (with $0 = N$). For example with $N=3$

$P = \begin{pmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{pmatrix}.$

Thus $P^N = I$. I want to invert the above equation to find $ S = F^{-1}(A) $ which I am sure can be done.

Does anyone have any suggestions for how to do this most effectively?

1 Answers 1

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Why are you "sure" that $F$ has an inverse? In fact, when $N=3$, $F$ has no inverse. To see this, note that $P^{-1}=P^2$ and $P^3=I$. So, if we put $S=P-I$, we have $ \begin{eqnarray} F(S) &=& 2S + P^{-1}S + 2SP + PSP\\ &=& 2(P-I) + P^{-1}(P-I) + 2(P-I)P + P(P-I)P\\ &=& (2P-2I) + (I-P^{-1}) + (2P^2-2P) + (P^3-P^2)\\ &=& (2P-2I) + (I-P^2) + (2P^2-2P) + (I-P^2) = 0. \end{eqnarray} $ So $F(S)=0$ for some nonzero $S$. Hence $F$ has no inverse.

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    I am interested in the positive (non-negative) matrices not the positive definite ones, however I believe that doesn't change the conclusions of the above comment as (like you said) for positive $S$, $S + \alpha uu^{T}$ will still be positive for sufficiently small $\alpha$.2011-09-09