Let $A$ and $B$ be two square real positive (all entries are positive) matrices that differ only in the first row. Let $\lambda_A$ and $\lambda_B$ be the maximal real eigenvalues of $A$ and $B$, respectively. Let $\lambda_*$ be the maximal real eigenvalue of the matrix $(A+B)/2$. It is easy to see that $\lambda_* \leq \max\{\lambda_A,\lambda_B\}$. I would like to know whether $\lambda_* \geq \min\{\lambda_A,\lambda_B\}$.

  • $\begingroup$ Could you add a few more details concerning the inequality "$\lambda_* \le \max\{\lambda_A,\lambda_B\}$"? I'm afraid I don't find it that easy to see... ;-) $\endgroup$ – Jochen Glueck Jun 23 '19 at 9:00
  • $\begingroup$ Additional remark: for arbitrary positive matrices $A,B$ (i.e. without the condition that $A$ and $B$ differ only in the first row) the inequality $\lambda_* \le \max \{\lambda_A, \lambda_B\}$ fails, in general. $\endgroup$ – Jochen Glueck Jun 23 '19 at 9:29
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    $\begingroup$ I just had a quick go with random $2\times 2$ matrices with entries drawn from $]0,1[$, trying out $10^9 $ cases. Without the condition that $A$ and $B$ only differ in the first row, 12% of cases had $\lambda_{*} $ outside $[\lambda_{A} ,\lambda_{B} ]$, with that condition, $\lambda_{*} $ was always inside the interval. $\endgroup$ – Michael Engelhardt Jun 23 '19 at 15:35
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    $\begingroup$ Jochen: the characteristic polynomial $f_A(\lambda) = det(\lambda I - A)$ of $A$ and the characteristic polynomial $f_B$ of $B$ are both positive for $\lambda$ larger than both $\lambda_A$ and $\lambda_B$, since the dominant term in $det(\lambda I - A)$ is $\lambda^n$ (and the maximal roots for these polynomials are $\lambda_A$ and $\lambda_B$, respectively). Since the matrices $A$ and $B$ differ only in one row, the characteristic polynomial $f_*$ of $(A+B)/2$ is $(f_A+f_B)/2$. Consequently, for every $\lambda$ larger than both $\lambda_A$ and $\lambda_B$, $f_*(\lambda)$ is positive. $\endgroup$ – Eilon Jun 24 '19 at 10:46
  • $\begingroup$ @Eilon: Thanks a lot; this is a very nice argument! $\endgroup$ – Jochen Glueck Jun 24 '19 at 19:17

Let $u$ be an eigenvector of $M = (A+B)/2$ for $\lambda_*$. By Perron-Frobenius we can choose $u \ge 0$. Now if $e_j$ is the $j$'th standard unit vector, $e_j^T A = e_j^T B$ for $j > 1$. Thus for $j > 1$, $e_j^T A u = e_j^T B u = \lambda_* u_j$. On the other hand, $e_1^T M u = \lambda_* u_1$ implies that $\max(e_1^T A u, e_1^T B u) \ge \lambda_* u_1$ and $\min(e_1^T A u, e_1^T B u) \le \lambda_* u_1$. WLOG $e_1^T A u \ge \lambda_* u_1$ and $e_1^T B u \le \lambda_* u_1$. Thus $v^T A u \ge \lambda_* v^T u$ and $v^T B u \le \lambda_* v^T u$ for any nonnegative vector $v$. In particular, this is true for the Perron eigenvector $v_A$ of $A^T$ and the Perron eigenvector $v_B$ of $B^T$. Thus $\lambda_A v_A^T u \ge \lambda_* v_A^T u$ and $\lambda_B v_B^T u \le \lambda_* v_B^T u$. Since the matrices have strictly positive entries, so do the Perron eigenvectors, and thus $\lambda_A \ge \lambda_*$ and $\lambda_B \le \lambda_*$.

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  • $\begingroup$ Thanks! This is very helpful. $\endgroup$ – Eilon Jun 24 '19 at 10:51
  • $\begingroup$ Suppose now that $\lambda_A > \lambda_B$. Is it true that $\lambda_* \in (\lambda_B,\lambda_A)$, or can it be that $\lambda_* = \lambda_B$ (or $\lambda_*=\lambda_A)$? $\endgroup$ – Eilon Jun 24 '19 at 13:23
  • $\begingroup$ Yes. Let $M(t) = t A + (1-t) B$, which has all entries strictly positive for $0 \le t \le 1$. The Perron eigenvalue is an analytic function $\lambda(t)$ of $t$ in a neighbourhood of $[0,1]$, with $\lambda_B = \lambda(0)$, $\lambda_A = \lambda(1)$ and $\lambda_* = \lambda(1/2)$. As in my answer, $\lambda(s)$ is between $\lambda(r)$ and $\lambda(t)$ when $s$ is between $t$ and $r$. Thus if $\lambda(1/2) = \lambda(0)$ we would have $\lambda(t) = \lambda_0$ for all $t \in [0,1/2]$, but then by analyticity $\lambda(t)$ is constant for all $t \in [0,1]$, making $\lambda_A = \lambda_B$. $\endgroup$ – Robert Israel Jun 24 '19 at 19:07
  • $\begingroup$ Thanks, Robert. In fact, I realized that your earlier proof delivers the strict monotonicity using Perron Frobenius Theorem: the vector $v_A$ is positive, hence by your argument above the inequality $\lambda_B v_B^T u < \lambda_* v_B^T u$ is strict. $\endgroup$ – Eilon Jun 24 '19 at 19:38

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