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I have a matrice in the form :

$$M = \begin{pmatrix} A & 0 & 0 \\\ B & A & 0 \\\ C & D & A \end{pmatrix} $$

where $A,B,C,D$ are diagonalizable square matrice and I want to determine

$$M^{\infty}:=\lim_{n\rightarrow \infty} M^n$$

in function of A,B,C,D.

Thank you for your help !

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I guess you are dealing with matrices with real or complex entries. First, we need a condition on the eigenvalues of $A$ in order to ensure $\lim_{p\to\infty}A^p$ exists in $\mathcal M_n(\Bbb C)$, as $A^p$ will be in the top-left corner of $M^p$. –  Davide Giraudo Nov 8 '12 at 16:05
    
Thank you for your answer ! The matrices have real entries. I know that $A$ has $1$ as eigenvalue and the others eigenvalues have a module strictly less than 1. –  Christophe Nov 8 '12 at 16:10
2  
You could include this in the OP. Do we have additional information, for example about the other matrices? –  Davide Giraudo Nov 8 '12 at 16:24

1 Answer 1

In general the limit will not exist. For example, the $(2,1)$ block of $M^n$ is $B_n = \sum_{j=1}^n A^{j-1} B A^{n-j}$. By taking a suitable basis, we may assume $A$ is diagonal. Under the assumption Christophe gave in a comment, that $1$ is an eigenvalue of $A$ and the other eigenvalues have absolute value $<1$, we can write $A = \pmatrix{I & 0\cr 0 & E\cr}$ where $E^n \to 0$ as $n \to \infty$. If $B$ has the corresponding block structure $\pmatrix{\alpha & \beta \cr \gamma & \delta\cr}$, then a necessary and sufficient condition for $B_n$ to have a limit as $n \to \infty$ is $\alpha = 0$. The limit of $B_n$ is then $\pmatrix{0 & \beta (I - E)^{-1}\cr (I-E)^{-1} \gamma & 0\cr}$. Similarly for the $(3,2)$ block with $B$ replaced by $D$. The requirements for the $(3,3)$ block to have a limit seem to be more complicated.

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