Is the spectral radius of $DA$ less then the one of $A$? Is the spectral radius of $DA$ less than the spectral radius of $A$ when $D$ is diagonal where all diagonal entries are nonnegative and less than 1?
This is true when $A$ is normal, since
$$
\rho(DA) \le \|DA\|\le \|D\| \| A\| \le \|A\| = \rho(A)
$$
My guess is that it is false in general.

Notice that it is enough to prove $\|(DA)^k\|\le \|A^k\|$ definitively in $k$.

If we let $D$ have negative values, then it would imply that any sign change in any row does not change the spectral radius, that is preposterous.
 A: This is false for every $n\ge2$. Pick any two vectors $u$ and $v$ such that $u_iv_i<0<u_jv_j$ for some $i\ne j$. Let $D=\operatorname{diag}(\operatorname{sign}(u_1v_1),\ldots,\operatorname{sign}(u_nv_n))$. Then
$$
v^TDu=\sum_i\operatorname{sign}(u_iv_i)u_iv_i=\sum_i|u_iv_i|>\left|\sum_iu_iv_i\right|=|v^Tu|.
$$
Therefore, when $A=uv^T$, we have  $\rho(DA)=|v^TDu|=v^TDu>|v^Tu|=\rho(A)$. By the continuity of spectral radius, we may reduce the diagonal entries of $D$ and perturb $A=uv^T$ to obtain other counterexamples such that $|d_{ii}|$ can be smaller than $1$ and $\operatorname{rank}(A)$ can be any number ranging from $1$ to $n$.
However, it is true that $\rho(DA)\le\rho(A)$ when we also have $A\ge0$ entrywise. This is because $(DA)^k\le A^k$ entrywise for every positive integer $k$, so that $\rho(DA)=\lim_{k\to\infty}\|(DA)^k\|_1^{1/k}\le\lim_{k\to\infty}\|(A)^k\|_1^{1/k}=\rho(A)$ by Gelfand's formula.
A: To add a counterexample with $D\ge 0$,
$$
A = \begin{pmatrix}
-1 & -1\\
1 & 1
\end{pmatrix}
$$
we have that $\rho(A) = 0$, but if $D$ is not a multiple of identity, we have $Trace(DA)\ne 0$ and so $\rho(DA)>0$.
