Suppose I have a square matrix $A$ that only has non-negative real entries and is not symmetric and not primitive either. It has no "special" structure we could exploit. I know that the spectral radius $\rho(A) <1$ (i.e. the maximum of the absolute eigenvalues of $A$ is smaller than $1$).

I now construct matrix $B$ by setting some of the entries of $A$ to zero (not specified how many or where they are). So we know the elementwise property that $A_{ij} \geq B_{ij} \geq 0$ for each $i,j$. We can also write the matrices as $A = B + C$ where the matrix $C$ contains the "deleted" values.

Is it possible to prove that if $\rho(A) <1$ then also $\rho(B) <1$?

Related postings:

  • When we have symmetric matrices we can use this.
  • For primitive matrices there is a handwavy argument here.
  • we can work with the upper bound of the row sum, but this did not give me the right result as it just provides an upper bound.
  • 4
    $\begingroup$ Within the cone of entry-wise positive matrices, the spectral radius is monotonous. This is a part of Perron-Frobeius theory. Thus $\rho(B)\le\rho(A)$, which solves your question. Have a look to my book Matrices (Springer-Verlag GTM 216) for instance. $\endgroup$ Nov 5, 2021 at 10:05
  • 1
    $\begingroup$ Incredible, thank you very much! I used Lemma 12 on page 153 of your book as a reference. $\endgroup$
    – blldt
    Nov 5, 2021 at 10:47


Browse other questions tagged .