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Let $D$ be a real diagonal matrix $D=\operatorname{diag}(a_1,a_2,\ldots,a_n)$ with $a_1\le a_2\le\ldots\le a_n$. Assume that at least one of the $a_i$ is positive. Let $P$ be an irreducible, real, row-stochastic matrix (all entries between 0 and 1, the sum of the elements of every row is 1), with all its eigenvalues real. Let $k>0$ be a real number. I wish to prove that $D+kP$ has a positive eigenvalue. Do you think this is true?

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Edit: Yes, it's true. Actually, the assumptions that $P$ is irreducible and row-stochastic can be removed. All we need is only that $P$ is (entrywise) nonnegative.

Since $P$ is nonnegative, so is $D-a_1I+kP$. By Perron-Frobenius theorem, the spectral radius $\rho(D-a_1I+kP)$ is an eigenvalue of $D-a_1I+kP$. Thus $\lambda=a_1+\rho(D-a_1I+kP)$ is a real eigenvalue of $D+kP$. It remains to show that $\lambda>0$.

As $D-a_1I+kP$ is entrywise bounded below by the nonnegative matrix $D-a_1I$, we see that $(D-a_1I+kP)^m$ is entrywise bounded below by $(D-a_1I)^m$ for every positive integer $m$. By using Gelfand's formula with Frobenius norm, it follows that $\rho(D-a_1I+kP)\ge\rho(D-a_1I)=a_n-a_1$. Therefore $\lambda=a_1+\rho(D+kP-a_1I)\ge a_n>0$.

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  • $\begingroup$ A clever and very elegant answer. Thank you! $\endgroup$ – user67490 Mar 20 '13 at 11:03
  • $\begingroup$ Very beautiful and elaborated answer. $\endgroup$ – user67692 Mar 20 '13 at 17:28
  • $\begingroup$ @user67490 You can also thank user1551 by voting up his answer. $\endgroup$ – user940 Apr 18 '13 at 20:35
  • $\begingroup$ @user67692 Beautiful enough to warrant an upvote, don't you think? $\endgroup$ – user940 Apr 18 '13 at 20:36

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