Spectral radius of matrix after column scaling

Question

Given an $nxn$ real matrix $A$ and a diagonal matrix $D=(d_1,\ldots,d_n)$, with each $d_i<1$. I would like to show that the spectral radius of $A\cdot D$ strictly decreases. That is, $\rho(A\cdot D) < \rho(A)$. Does this hold in general? If not, does it hold when $A$ is symmetric and has spectral radius $1$ ($\rho(A)=1$)? In fact, the $A$ for which I need the result is the Laplacian of the complete graph $K_n$.

Answer 1

Presumably $A\ne0$ and $0\le d_i<1$ for each $i$. Then $\rho(AD)<\rho(A)$. This is true not only when $A$ is Laplacian, but also for every nonzero Hermitian matrix $A$, for, if $d=\frac{1+\max_i\sqrt{d_i}}2$, then $\|x\|>d\|x\|>\max_id_i\|x\|\ge\|Dx\|$ for every nonzero $x$ and hence $$ \begin{align*} \rho(A) >d^2\rho(A) &=d^2\max_{\|x\|\le1}\|Ax\|\\ &=d\max_{\|y\|\le d}\|Ay\|\\ &\ge d\max_{\|z\|\le1}\|AD^{1/2}z\|\\ &\ge\max_{\|z\|\le1}\|D^{1/2}AD^{1/2}z\|\\ &=\rho(D^{1/2}AD^{1/2})\\ &=\rho(AD). \end{align*} $$ If $A$ is not Hermitian, the inequality does not hold in general. E.g. when $$ A=\pmatrix{1&1\\ -1&-1} $$ and $D$ has two distinct positive entries, $\rho(A)=0<\rho(AD)$ because $A$ is nilpotent but $AD$ is not.