Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $\rho(A) = \frac{\sqrt{481}+11}{40}<1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.