# Largest eigenvalue of matrix A smaller than 1, what about B when A=B+C? [closed]

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.
• 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. Nov 5, 2021 at 10:05
• Incredible, thank you very much! I used Lemma 12 on page 153 of your book as a reference. Nov 5, 2021 at 10:47