Suppose that $M$ is an $n\times n$ matrix with $\rho(M)<1$ (i.e. its maximum absolute eigenvalue is less than $1$).
Is the following statement then true?
If $\forall t\in\mathbb{N}$, $D_t$ is an $n \times n$ diagonal matrix with $D_{t,ii}\in[-1,1]$ ($\forall i\in {1,\dots,n}$), then: $$\lim_{t\rightarrow\infty}\prod_{s=1}^t{D_s M}=0.$$
My intuition says it ought to be, but I may be wrong. My thinking is that without the $D_t$s, this would just be the standard result on the spectral radius, and multiplying by diagonal matrices with values in the unit circle ought (?) not to change things.
If the result does indeed hold, then I'd also be interested in the converse (i.e. does the highlighted statement imply that $\rho(M)<1$?).
No. As a simple example, consider $$ M=\pmatrix{1 & 1/2 \\ -1/2 & 0}. $$ with $\rho(M)=1/2<1$ and the simple choice $$ D_s=D=\pmatrix{1&0\\0&-1} $$ for all $s$. Then $\rho(DM)=(2^{1/2}+1)/2>1$ so $(DM)^t$ diverges with $t\rightarrow\infty$.
It is not generally true even if the diagonal entries of $D$'s are in $(0,1)$. E.g., with $$ M=\frac{1}{2}\pmatrix{17/2 & 8 \\ -8 & -15/2}, \quad D=\pmatrix{1/2 & 0 \\ 0 & 3/4} $$ $\rho(M)=1/2$ but $\rho(DM)=(97^{1/2}+11)\approx 1.3>1$.
However, it would be true if $M$ and $D_t$'s were nonnegative matrices. One might have then $\prod_{s=1}^t(D_sM)\leq M^t$ which converges for $t\rightarrow\infty$.