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:

Matrices(Springer-Verlag GTM 216) for instance. $\endgroup$