Eigenvalue multiplicity of a product matrix

Question

Let $A_{m}\cdot A_{m-1}\cdots\cdot A_{1}$ be a left product of $m$ diagonalizable $n$-dimensional distinct Markov matrices with strictly positive entries. Assume that each matrix has $n$ distinct eigenvalues. Let $\lambda^{k}_j$ the $k^{th}$ eigenvalue of matrix $A_j$. Assume without loss of generality that $\lambda^{k+1}_j > \lambda^{k}_j$ for $k=1,\ldots n-1$ for every matrix $j=1,\ldots, m$. The determinant of the matrix $A= A_{m}\cdot A_{m-1}\cdots\cdot A_{1}$ can be written as: \begin{equation} Det\left(A\right)=Det\left(A_m\right)\cdot Det\left(A_{m-1}\right)\cdot\ldots\cdot Det\left(A_1\right)=\prod_{j=1}^{m}\lambda_{j}^1\cdot\prod_{j=1}^{m}\lambda_{j}^2\cdot\ldots\cdot\prod_{j=1}^{m}\lambda_{j}^n \end{equation} As the number of matrices increases, the determinant of the matrix $A= A_{m}\cdot A_{m-1}\cdots\cdot A_{1}$ goes to zero as soon as $\prod_{j=1}^{m} \lambda_{j}^1$ goes to zero. What can be said about the algebraic and geometric multiplicity of the zero eigenvalue of the product matrix $A$? Assume that also $\prod_{j=1}^{m} \lambda_{j}^2$ goes to zero. Again, what can be said about the algebraic and geometric multiplicity of the zero eigenvalue of the product matrix $A$?