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I want to prove (or disprove) the following statement:

If $A$ is a square matrix with non-negative elements that has spectral radius less then $1$, then any matrix obtained from $A$ by arbitrarily changing the sign of the elements has the same property.

This problem appeared recently when studying the convergence of some matrices and I would like to believe that is true.

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up vote 5 down vote accepted

This is true. Recall that if $\|A\|$ is the operator norm of an $n\times n$ matrix $A$ and $\|A\|_\infty$ is the maximum absolute value of the entries of $A$, then,

$$\|A\|_\infty \le \|A\| \le n\|A\|_\infty.$$

If $B$ is an arbitrarily signed version of the non-negative matrix $A$, the triangle inequality shows that $\|B^k\|_\infty \le \|A^k\|_\infty$ for any $k\ge 1$. By Gelfand's formula we have: $$ \rho(B) = \lim_{k\to\infty} \|B^k\|^{1/k} \le \lim_{k\to\infty} (n\|B^k\|_\infty)^{1/k} \le \lim_{k\to\infty} (n\|A^k\|_\infty)^{1/k} \le \lim_{k\to\infty} (n\|A^k\|)^{1/k} = \rho(A).$$

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