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Is it true that if a matrix has nonnegative elements and spectral radius less than $1$, than the sum of its elements on each row (and column) is less than $1$?

Edit: What if the matrix has positive elements?

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The reverse implication is correct: if the sum of row (column) elements of $A$ is less than 1 you have $\|A\|_1 < 1$ resp. $\|A\|_{\infty} < 1$ and there is the rule that $\rho(A) \le \|A\|$ for all operator norms. –  Cocopuffs Jul 20 '12 at 15:23

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up vote 3 down vote accepted

$A=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$. The spectral radius is $0$, but a row sum is $1$.

A counterexample with all entries strictly positive follows immediately from continuity of eigenvalues.

For an explicit example, try $A=\begin{bmatrix}\frac{1}{10} & 1 \\ \frac{1}{10} & \frac{1}{10} \end{bmatrix}$. It is straightforward to check that $\rho(A) = \frac{\sqrt{10}+1}{10}<1$.

For a symmetric example, take $A=\begin{bmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{20} \end{bmatrix}$. This has $\rho(A) = \frac{\sqrt{481}+11}{40}<1$.

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Of course (silly me). What if the matrix has positive elements? (I edited the post to include this question, too) –  digital-Ink Jul 20 '12 at 15:15

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