# Convergence of powers of matrices in Jordan Canonical Form (Jordan Normal Form)

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

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).

• That makes so much sense. Thank you so much! – user3424575 May 10 at 0:42