the convergence of the infinite product of matrices composed by matrces with the spectral radius less than 1

Question

Suppose the two matrices $\mathbf{A}$ and $\mathbf{B}$ have the spectral radius less than $1$. Do the infinite product $\mathbf{C}\left( 1 \right)\mathbf{C}\left( 2 \right)\cdots $ , where each $\mathbf{C}\left( t \right),\ t=1,2,\cdots $ is $\mathbf{A}$ or $\mathbf{B}$, converge to $\mathbf{o}$ ? I tried to prove it using Jordan canonical forms or matrix norms, but I couldn't. Please tell me how to prove this claim or show the counterexample.

Answer 1

For a counterexample, let $A,B$ be given by \begin{align*} A&= \pmatrix{ 0&\frac{1}{4}\\ -1&1\\ } \\[4pt] B&= \pmatrix{ 0&-\frac{1}{4}\\ 1&1 } \end{align*} Then the eigenvalues of $A$ are ${\large{\frac{1}{2}}},{\large{\frac{1}{2}}}$, and the eigenvalues of $B$ are also ${\large{\frac{1}{2}}},{\large{\frac{1}{2}}}$, so each of $A,B$ has spectral radius less than $1$.

Computing $AB$, we get $$ AB = \pmatrix{ \frac{1}{4}&\frac{1}{4}\\ 1&\frac{5}{4}\\ } $$ which has eigenvalues $$ \frac{3}{4}+\frac{\sqrt{2}}{2},\;\frac{3}{4}-\frac{\sqrt{2}}{2} $$ so $AB$ has spectral radius greater than $1$.

It follows that $(AB)^n$ doesn't approach $0$ as $n$ approaches infinity.