Infinite Product of 2 Matrices

Question

Let us assume that I have the $n \times n$ matrices $A$ and $B$ where $n$ is finite.

I construct the matrix infinite matrix product $P = \lim_{N \rightarrow \infty}\prod_{i=1}^{N}X_{i}$ and, for each $i$, $X_{i}$ is either $A$ or $B$. In total, $X_{i} = A$ and $X_{i} = B$ for $N_{A}$ and $N_{B}$ instances respectively, with $N_{A} + N_{B} = N$. Let us also say that the ratio of $N_{A}$ and $N_{B}$ remains finite and non-zero (which requires $N_{A},N_{B} \rightarrow \infty$ as $N \rightarrow \infty$).

Let $\sigma_{1}$ denote the (possibly degenerate) largest singular value of $P$ and $\sigma_{2}$ denote the next largest singular value of $P$ that is distinct from $\sigma_{1}$, i.e. $\sigma_{1} > \sigma_{2}$.

Is it true that $\frac{\sigma_{2}}{\sigma_{1}} = 0$ always? I.e. $\sigma_{\rm 1}$ is infinitely larger than any of the other distinct singular values.

Edit: Simple example: $B$ is the identity matrix and $A$ is normal. The singular values of $P$ are therefore $(\sigma_{i})^{N_{A}}$ where $\sigma_{i}$ are the singular values of $A$. Clearly the largest singular value is infinitely larger than the rest as $N_{A} \rightarrow \infty$.

Answer 1

OP asks

Let $\sigma_{1}$ denote the (possibly degenerate) largest singular value of $P$ and $\sigma_{2}$ denote the next largest singular value of $P$ that is distinct from $\sigma_{1}$, i.e. $\sigma_{1} \gt \sigma_{2}$. Is it true that $\frac{\sigma_{2}}{\sigma_{1}} = 0$ always?

This is not true. Consider $B:=I_n$ and
$A:= \left[\begin{array}{cc}\frac{1}{m}\mathbf {1_m1_m}^T & \mathbf 0 \\\mathbf 0 & \mathbf \pi\mathbf 1_{n-m}^T \end{array}\right]$
where $\mathbf 1_d$ is the vector of all ones in $\mathbb R^d$ and $\pi$ denotes any non-uniform probability vector in $\mathbb R^{n-m}$

$A$ and $B$ are both idempotent and commute, thus
$P=A$ and $\text{rank}(P)=2$ where $\sigma_1 \gt 1 = \sigma_2$
But $\sigma_1$ is finite and in fact $\sigma_1=\Big\Vert \mathbf \pi\mathbf 1_{n-m}^T\Big \Vert_F= \sqrt{n-m}\cdot \Big \Vert \pi\Big\Vert_2\gt 1$
(e.g. check Cauchy-Schwarz to verify that the inequality is strict)

note: OP also states

Edit: Simple example: B is the identity matrix. The singular values of $P$ are therefore $(\sigma_{i})^{N_{A}}$ where $\sigma_{i}$ are the singular values of $A$. Clearly the largest singular value is infinitely larger than the rest as $N_{A} \rightarrow \infty$.

This is false. Singular values do not in general multiply for non-normal matrices. The above idempotent $A$ is a counterexample to this assertion.