# 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.

• ok, I have a $2\times 2$ counterexample.. Jul 18 '20 at 11:51

This is false for every $$n\ge2$$. Pick any two vectors $$u$$ and $$v$$ such that $$u_iv_i<0 for some $$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)$$ by Gelfand's formula.
To add a counterexample with $$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$$.