2
$\begingroup$

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 $\lambda_* \geq \min\{\lambda_A,\lambda_B\}$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
6
$\begingroup$

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 $\lambda_B \le \lambda_*$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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