Let $M$ be a irreducible, symmetric matrix with some negative entries such that $M^k>0$ for some $k>k_o$ and $\sum_j m_{ij}=1$ with $m_{ij} \in \Re$, and spectral radius $\rho(M)=1$. After multiplying it with a diagonal matrix $D={\rm diag}(d_{ii})$ where $0<d_{ii}\leq 1$ with at least one $d_{ii}<1$. Is there an easy way to show $\rho(DM)<1$? I know that this result is true as stated.
Similar questions have been asked for matrices $M$ but only for nonnegative entries, e.g., Substochastic matrix spectral radius.
Best Answer
I use OP's definition that $M^k$ is a positive matrix for all $k$ large enough. This means $M$ has a single eigenvalue on the unit circle that is simple and $=1$ and an associated Perron vector that is strictly positive (apply Perron theory to $M^k$ which thus has a single eigenvalue on unit circle $=1$ and this is simple, now work backwards).
WLOG suppose $d_n \in(0,1)$ Now using the operator 2 norm write
$\big \Vert DM\big\Vert_2 = \max_{\mathbf x , \mathbf y \in S^{n-1}}\big\vert \mathbf x^TD M\mathbf y\big\vert \leq \max_{\mathbf x , \mathbf y \in S^{n-1}}\big\Vert D\mathbf x\big \Vert_2\big\Vert M\mathbf y\big \Vert_2\leq 1 \cdot 1$
where the first inequality is Cauchy-Schwarz which is met with equality iff $ \alpha \cdot D\mathbf x = M\mathbf y$ and the RHS is met with equality iff $ D\mathbf x =\mathbf x\implies x_n=0$ and $M\mathbf y = \mathbf y \implies \mathbf y$ is a positive vector. So the upper bound being met with equality implies $\alpha \cdot \mathbf x =\mathbf y$ where the left hand side has a zero in its $n$th component and the vector on the right is positive. This is impossible.
Conclude $\lambda_\text{max modulus}(DM)\leq \big \Vert DM\big\Vert_2\lt 1$