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.

  • 1
    $\begingroup$ How can a stochastic matrix have negative entries? $\endgroup$ May 9, 2023 at 10:50
  • $\begingroup$ @charmd It is true because of the property of irreducibility, the multiplication of row with a value less than 1 is propagated through out the matrix. The matrix $DM$ is a nilpotent matrix. I don't have a stringent proof for this. $\endgroup$
    – Desperado
    May 9, 2023 at 11:02
  • 2
    $\begingroup$ The question as stated is unclear. You have to define "eventually stochastic matrix" $\endgroup$
    – parsiad
    May 9, 2023 at 15:45
  • $\begingroup$ To be clear: does "eventually positive" actually mean positive or does it mean non-negative in entries? $\endgroup$ May 9, 2023 at 16:37
  • $\begingroup$ @Ahsan why did you delete your response to my above question? For this question to make sense it is vital that people see your prior response of $M^k$ having strictly positive entries for some $k$. Also your newest edit to the original post doesn't make sense as there is a single standard definition for a stochastic matrix and your allowance of negative entries violates this. The information in the version of the original post prior to this edit and your now deleted comments were both necessary to make it a complete post. $\endgroup$ May 10, 2023 at 16:19

1 Answer 1


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 $\lambda_\text{max modulus}(DM)\leq \big \Vert DM\big\Vert_2\lt 1$

  • $\begingroup$ Hi, you assume $d_n \in (0,\,1)$, but $d_n$ can be 1. If $d_n \in (0,\,1)$ is used, then proof is a lot simpler where $\rho(DM) \leq \rho(D)\rho(M)$ with $\rho(M)=1$, $\rho(D)<1$ and where $\rho(.)$ is the spectral radius of a matrix. $\endgroup$
    – Desperado
    May 9, 2023 at 18:56
  • 1
    $\begingroup$ No. By $d_n$ I meant the nth component of $D$ -- this would be $d_{n,n}$ in your notation. The spectral radius of $D$ is assumed to be $1$. If the Without Loss of Generality is confusing, you can replace that $n$ with "$r$ for some $r\in \big\{1,2,\dots,n\big\}$" in the two times it comes up. $\endgroup$ May 9, 2023 at 19:14
  • $\begingroup$ Hi, sorry I have one more question. I know I said M is symmetric but It looks like your answer is for any $M$ with a positive left eigenvector. But I have an example where $M=\begin{bmatrix} 0.6683 & -0.5264 & 0.2627 & 0.5954\\ 0.0580 & 0 & 0.3073 & 0.6347\\ 0.1093 & 0.1965 & 0.9058 & -0.2115\\ 0.5888 & -0.4422 & 0.1360 & 0.7173 \end{bmatrix}$ and $D={\rm diag}([0.9413,\, 0.5038,\, 0.6997,\, 0.9123])$ result in $\rho(DM)>1$. So what is the catch here. is there any restriction on $M$ $\endgroup$
    – Desperado
    May 9, 2023 at 20:55
  • $\begingroup$ In your proof, I think you need the simplicity of the eigenvalue $1$ to justify that $\|My\|_2=1\Longrightarrow My=y$, because in general, a singular vector of a symmetric matrix is not necessarily an eigenvector. It seems that the proof can be made simpler if instead of $\|DM\|_2$ you consider $v^TDMv$ for an eigenvector $v$ of $DM$ corresponding to a dominant eigenvalue. $\endgroup$
    – user1551
    May 9, 2023 at 21:20
  • 1
    $\begingroup$ Yes. As I said two weeks ago in the above: we have $D\mathbf x = \mathbf x$ implies $x_n=0$ since $D$ has some diagonal component $\in (0,1)$ and I had assumed Without Loss of Generality that it was the $n$th component. In my original comment to this answer I further explained how to interpret the Without Loss of Generality assumption in case you were unfamiliar. Please re-visit that comment. $\endgroup$ May 24, 2023 at 16:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .