1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ ok, I have a $2\times 2$ counterexample.. $\endgroup$
    – Exodd
    Jul 18 '20 at 11:51
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\rho(DA)>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.