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Given a square matrix $A$ with certain properties ($A$ is diagonalizable, $\rho(A) = 1$ and each column of $A$ adds up to unity) and two square matrices $M$ and $N$ of rank $1$ (both matrices are products of a fixed column vector and an arbitrary row vector), is it possible to find a closed form expression for the infinite sum

$$S = \sum_{k=0}^\infty A^k M A^k N A^k$$

which I know to be convergent (but I do not know what it converges to)?

For matrices $B$ with $\rho(B) < 1$ it might be useful to rewrite

$$\sum_{k=0}^\infty B^k = (I-B)^{-1},$$

but even then the problem would not be solved since

$$\sum_{k=0}^\infty B^k M B^k N B^k \neq (I-B)^{-1} M (I-B)^{-1} N (I-B)^{-1}.$$

[Edit] Since $A$ can be diagonalized as $A = V \Lambda V^{-1}$ and hence $A^k = V \Lambda^k V^{-1}$ (see also my comment below), the above sum can be rewritten to

$$S = V \left( \sum_{k=0}^\infty \Lambda^k \bar{M} \Lambda^k \bar{N} \Lambda^k \right) V^{-1}$$

with $\bar{M} = V^{-1} M V$ and likewise $\bar{N} = V^{-1} N V$. Any suggestions on how to proceed?

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What is $\rho(A)$? –  user7530 Apr 15 '14 at 16:05
@user7530 It is the spectral radius of the matrix $A$, its maximum eigenvalue. –  Ailurus Apr 15 '14 at 16:42
I would try first with a diagonal matrix $A$, to see if at least in this case a closed-form formula exist. –  Matemáticos Chibchas Apr 15 '14 at 22:11
@MatemáticosChibchas Ok, but $A$ can be diagonalized (e.g. $A = V \Lambda V^{-1}$ such that $A^k = V \Lambda^k V^{-1}$), which results in $V \left( \sum_{k=0}^\infty \Lambda^k \bar{M} \Lambda^k \bar{N} \Lambda^k \right) V^{-1}$. –  Ailurus Apr 16 '14 at 9:35

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