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I've actually been stuck on this for a bit while studying for an exam, so would appreciate any help.

The problem involves testing whether $\lim\limits_{m \to \infty}$ $A^m$ exists.

From lecture notes, I know that the limit above exists for a matrix A when A has no eigenvalue $\lambda > \lvert 1\rvert$. I also know this limit exists if and only if the limit exists for the Jordan canonical form for the matrix. I was given the following theorem in class with respect to matrices in JCF, however I'm unable to apply it (perhaps poor understanding of Jordan canonical matrices):

"Let A be an nxn matrix. Then $\lim\limits_{m \to \infty}$ $A^m$ only exists if:

Matrix A has an eigenvalue $ \lvert \lambda\rvert = 1$, then $\lambda = 1$ and all the blocks in the Jordan normal form $J(A)$ of the form $J_k(1)$ must have $k = 1$.

I am unable to understand how to apply this to the following matrices to determine whether $\lim\limits_{m \to \infty}$ $A^m$ exists.

1) $J(A_1)$ = $$ \begin{pmatrix} 1/2 & 1 & 0 & 0 & 0 \\ 0 & 1/2 & 1 & 0 & 0 \\ 0 & 0 & 1/2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

2) $J(A_2)$ = $$ \begin{pmatrix} 1/2 & 1 & 0 & 0 & 0 \\ 0 & 1/2 & 1 & 0 & 0 \\ 0 & 0 & 1/2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$

3) $J(A_3)$: $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$

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For the first matrix, the Jordan blocks are:

$$J(A_1)= \begin{pmatrix} \color{red}{1/2} & \color{red}1 & \color{red}0 & 0 & 0 \\ \color{red}0 & \color{red}{1/2} & \color{red}1 & 0 & 0 \\ \color{red}0 & \color{red}0 & \color{red}{1/2} & 0 & 0 \\ 0 & 0 & 0 & \color{blue}1 & 0 \\ 0 & 0 & 0 & 0 & \color{brown}1 \\ \end{pmatrix}. $$ In particular, there is a $3 \times 3$ block with eigenvalue $1/2$ (convergent, since $|1/2| < 1$) and two $1 \times 1$ blocks with eigenvalue $1$ (convergent even though $|1| = 1$, since the blocks are only $1 \times 1$). So, the powers of $A_1$ are convergent.

For the second matrix, the Jordan blocks are:

$$J(A_2) = \begin{pmatrix} \color{red}{1/2} & \color{red}1 & \color{red}0 & 0 & 0 \\ \color{red}0 & \color{red}{1/2} & \color{red}1 & 0 & 0 \\ \color{red}0 & \color{red}0 & \color{red}{1/2} & 0 & 0 \\ 0 & 0 & 0 & \color{blue}1 & \color{blue}1 \\ 0 & 0 & 0 & \color{blue}0 & \color{blue}1 \end{pmatrix}. $$

The $3 \times 3$ Jordan block is still convergent, but now the two $1 \times 1$ Jordan blocks have been replaced by a $2 \times 2$ Jordan block with eigenvalue $1$. Since there is a Jordan block corresponding to $1$ that is larger than $1 \times 1$, the powers of $A_2$ are divergent.

For the third matrix, the Jordan blocks are:

$$J(A_3) = \begin{pmatrix} \color{red}1 & 0 & 0 \\ 0 & \color{blue}{-1} & 0 \\ 0 & 0 & \color{brown}1 \\ \end{pmatrix}. $$ In this case, there are three $1 \times 1$ Jordan blocks. The two blocks with eigenvalue $1$ are not a problem, but the one with eigenvalue $-1$ is an a problem (while $|-1| = 1$, we don't have $-1 = 1$). So, the powers of $A_3$ are not convergent (it's not hard to compute $J(A_3)^n$ and verify that it alternates, rather than converging).

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    $\begingroup$ That makes so much sense. Thank you so much! $\endgroup$ – user3424575 May 10 at 0:42

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