Spectral radius of a matrix.

Question

Suppose I have a matrix $A$ which has the following property $\rho(A)\leq \rho(A_{\rm sym})$, is it true that $\rho(DAD)\leq \rho(DA_{\rm sym}D)$ for a positive definite diagonal matrix $D$?

Answer 1

I suppose that $A_{\text{sym}}$ refers to the symmetric part of $A$, i.e., $A_{\text{sym}}=\frac12(A+A^T)$. Then the answer to your question is negative. It should be easy to generate a random counterexample by computer. E.g. $$ \begin{aligned} &A=\pmatrix{1&4\\ -2&4}, \quad A_{\text{sym}}=\pmatrix{1&1\\ 1&4}, \quad D=\pmatrix{1&0\\ 0&\frac12},\\ &DAD=\pmatrix{1&2\\ -1&1}, \quad DA_{\text{sym}}D=\pmatrix{1&\frac12\\ \frac12&1},\\ &\rho(A)=\sqrt{12}<\frac{\sqrt{12}+\sqrt{13}}{2}<\frac{5+\sqrt{13}}{2}=\rho(A_{\text{sym}}),\\ &\rho(DAD)=\sqrt{3}>\frac32=\rho(DA_{\text{sym}}D). \end{aligned} $$