# The maximal eigenvalue of average of positive matrices

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

• 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... ;-) Jun 23 '19 at 9:00
• 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. Jun 23 '19 at 9:29
• 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. Jun 23 '19 at 15:35
• 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. Jun 24 '19 at 10:46
• @Eilon: Thanks a lot; this is a very nice argument! 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 $$\lambda_B \le \lambda_*$$.
• 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 $\lambda_*=\lambda_A)$? Jun 24 '19 at 13:23
• 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 $\lambda_A = \lambda_B$. Jun 24 '19 at 19:07
• 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. Jun 24 '19 at 19:38