MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 !

share|cite|improve this question
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
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.