Eigenvector quadratic form inequality

Question

Given matrix $B\in\mathbb{R}_+^{n\times n}$ and scalar $\alpha \in \mathbb{R}_{+}$, let $A:=\alpha B+B^T/\alpha$. Note that $B$ and $A$ have nonnegative entries and that $\alpha$ controls degree of asymmetry of $A$. Let $v \in \mathbb{R}^n_+, \Vert v\Vert =1$ be the Perron eigenvector of $A$ that achieves largest (real-valued) eigenvalue. Numerical explorations suggest the following simple inequality: $$\vert u^* B u \vert \le v^T B v$$ whenever $u\in\mathbb{C}^n, \Vert u \Vert =1$ is an eigenvector of $A$. This is easy to prove for $\alpha=1$ (symmetric $A$) using Courant-Fischer. Any ideas for how to prove or disprove this for other $\alpha$?

[Note: This builds off another one of my MO question's. However, here it is formulated in a more general and self-contained setting.]

Answer 1

EDIT: Unfortunately, I just realized the $B$ matrix in the counterexample below is not positive! I still suspect this to be false, but constructing an example satisfying the necessary constraints will be slightly more subtle. I will think over it further.

Flawed counterexample: Let $1 < \alpha < \sqrt{3/2}$. Then setting $$B = \frac{1}{\alpha^2-1/\alpha^2}\left(\alpha \begin{bmatrix}6 & 7 & 2 \\ 3& 4 & 8 \\6 & 4 & 5 \end{bmatrix}- \frac{1}{\alpha}\begin{bmatrix}6 & 3 & 6 \\ 7& 4 & 4 \\2 & 8 & 5 \end{bmatrix} \right),$$ we will get that $$A = \alpha B + B^T/\alpha= \begin{bmatrix}6 & 7 & 2 \\ 3& 4 & 8 \\6 & 4 & 5 \end{bmatrix},$$ which has a Perron eigenvalue of $15$ corresponding to a normalized eigenvector $$v = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},$$ and also an eigenvalue of $3i$ corresponding to the normalized eigenvector $$u = \frac{1}{2\sqrt{6}}\begin{bmatrix} -1 + 3i \\ -1 - 3i \\ 2 \end{bmatrix}.$$ Computing, we see that $$|u^\ast B u| = \left|\frac{3i}{\alpha-1/\alpha} \right| = \frac{3}{\alpha - 1/\alpha}.$$ while $$v^T B v = \frac{15}{\alpha + 1/\alpha}.$$ Finally, we note that for all $\alpha$ such that $1 < \alpha < \sqrt{3/2}$, one can easily verify that $$\frac{15}{\alpha + 1/\alpha} < \frac{3}{\alpha - 1/\alpha}.$$

The key idea behind constructing this example is that, unless $\alpha = 1$, one can generally reconstruct $B$ from $A$, and then simply translate the problem back into one in terms of $A$. We see that the desired is equivalent to a bound on the size of the eigenvalues of $A$ relative to the Perron eigenvalue. This will be true automatically if the eigenvalue is real, but I expect one can construct counterexamples otherwise, as I tried and failed to do above. This is shown in the analysis below.

Analysis

Note that $$A + A^T = (\alpha + 1/\alpha)(B+B^T),$$ and $$A-A^T = (\alpha-1/\alpha)(B-B^T),$$ so we have that \begin{align*} B &= \frac{1}{2}\left(\frac{1}{\alpha+1/\alpha}(A+A^T)+\frac{1}{\alpha-1/\alpha}(A-A^T)\right) \\&= \frac{1}{2}\left(\frac{1}{\alpha+1/\alpha}+\frac{1}{\alpha-1/\alpha}\right)A + \frac{1}{2}\left(\frac{1}{\alpha+1/\alpha}-\frac{1}{\alpha-1/\alpha}\right)A^T. \end{align*} Some basic algebra shows $$\frac{1}{\alpha+1/\alpha}+\frac{1}{\alpha-1/\alpha} = \frac{2\alpha}{\alpha^2-1/\alpha^2} $$ and $$\frac{1}{\alpha+1/\alpha}-\frac{1}{\alpha-1/\alpha} = \frac{-2/\alpha}{\alpha^2-1/\alpha^2},$$ so $$B = \frac{1}{\alpha^2-1/\alpha^2}\left(\alpha A - A^T/\alpha \right),$$ and therefore that $$u^\ast B u = \frac{1}{\alpha^2-1/\alpha^2}\left(\alpha u^\ast A u - u^\ast A^T u/\alpha \right)$$ for any $u$ in $\mathbb{C}^n$. In particular, if $u$ is a unit eigenvector of $A$, then $u^\ast A u$ is the corresponding eigenvalue $\lambda$ and $u^\ast A^T u$ is its complex conjugate $\overline{\lambda}$ (to see this, note that $u^\ast A^T u$ is the conjugate transpose of the one-by-one matrix $u^\ast A u$, as $A$ has real entries). Therefore, for a unit eigenvector $u$ of $A$ with eigenvalue $\lambda$, we have $$|u^\ast B u| = \left|\frac{1}{\alpha^2-1/\alpha^2}(\alpha \lambda - \overline{\lambda}/\alpha) \right| = \frac{1}{\alpha^2-1/\alpha^2}\left | \alpha \lambda - \overline{\lambda}/\alpha \right|.$$

On the other hand, because $v$ is a real eigenvector of $A$ with eigenvalue $r = \rho(A)$, we have that $v^T B v = v^T B^T v$ and $v^T A v = v^T A^T v = r$ (by taking the one-by-one transpose), we have that $$v^T B v = \frac{1}{2}(v^T (B + B^T)v) = \frac{1}{2(\alpha + 1/\alpha)}(v^T(A + A^T)v) = \frac{r}{\alpha + 1/\alpha}.$$ Therefore, the desired inequality states that, for any eigenvalue $\lambda$ of $A$, $$\frac{1}{\alpha^2-1/\alpha^2}\left | \alpha \lambda - \overline{\lambda}/\alpha \right| \leq \frac{r}{\alpha + 1/\alpha},$$ which is equivalent to $$\frac{1}{\alpha-1/\alpha}\left | \alpha \lambda - \overline{\lambda}/\alpha \right| \leq r,$$ which is again equivalent to $$|\lambda| \frac{|\alpha - e^{-2i\theta} /\alpha|}{\alpha-1/\alpha} \leq r,$$ where $\theta$ is a complex argument of the eigenvalue $\lambda$. If $\theta = 0$ (i.e. the eigenvalue $\lambda$ is real), this automatically holds (as $r$ is strictly larger than all other eigenvalues in magnitude, and the factor on the left side is just one). However, if not, I expect we can force the factor on the left side to be larger than one.

Answer 2

Sorry for deleting and adding my non-answer. I got tricked by how simple looking the problem was but it turned out to not to be simple at all. After playing numerically just like yourself I am convinced this is true. I don't have a proof but I managed to gather some relevant data from the literature where the ideas in it could lead to a proof (I know this was not quite what you were looking for).

First of all the problem can be reformulated so it looks more natural. The coefficients $\alpha$ and $1/\alpha$ are making it confusing and they are not useful. Instead its is better to normalize the coefficients so they sum up to $1$. So you can ask: for a positive real matrix $B$, let's denote $tB+(1-t)B^T$ for $t\in [0,1]$ by $B_t$. Then for any any normalized eigenvector $u$ of $B_t$ does $u^*Bu$ get maximized when $u$ is the Perron eigenvector?

The benefit of this is that $B_t$ has a name in literature, it is called Levinger transformation. Furthermore assuming $t\neq 1/2$ (which is equivalent to $\alpha=1$ and you already know how to prove that), we can reformulate in terms of eigenvalues of $B_t$. A simple calculation and looking at the real and imaginary parts of $u^*Bu$ (similar to what the other answer has done) you get the following reformulation:

Assuming $t\in [0, 1]$ and $t\neq 1/2$, let's denote the spectral radius of $B_t$ by $\rho(B_t)$ (this is the Perron eigenvalue), then for any eigenvalue $\lambda = a+ bi$ where $a, b \in \mathbb{R}$ we have: $$a^2+(\frac{1}{1-2t})^2b^2 \leq \rho(B_t)^2$$

Possible approaches and some ideas from literature: You can define the function $f(t) = \rho(B_t)^2 - a^2-(\frac{1}{1-2t})^2b^2$. We want to prove $f(t) \geq 0$. This obviously holds for $t=0,1$ due to Perron-Frobenius. I suspect $f(t)$ might be increasing on the interval $[0, 1/2)$ and decreasing on $(1/2, 1]$. So checking the inequality for $0$ and $1$ is enough. For this you would need to calculate the derivative of $f$ and for that you will need to use perturbation theory to approximate the eigenvalues under small perturbation. You will probably need to assume that $B_t$ has distinct eigenvalue and this is an assumption you can make. Because it suffices to prove the inequality for a dense subset of matrices. Matrices with distinct eigenvalues that are real and positive are dense (you can use real Jordan form or real Schur decomposition to see this). So you can assume that $B$ has distinct eigenvalues. Furthermore when $B_t$ moves from $B$ to $B^T$ in at most finitely many points it can have non-distinct eigenvalues (you just need to look at the discriminant of the characteristic polynomial of $B_t$ and it can vanish in at most finitely many points). This means you can assume that you are working on an interval where $B_t$ has distinct eigenvalues and theorems from perturbation theory apply.

This kind of approach has been used by Levinger to prove that $\rho(B_t)$ is increasing on $[0, 1/2)$. You can take a look at Numerical range of matrices and Levinger's theorem.

I also found some interesting upper bounds on real and imaginary parts of eigenvalues of $B_t$ in The generalized Levinger transformation, Theorem 1. But I don't think it is sharp enough to prove your conjecture. But the ideas might be helpful.

Another reformulation of the problem: Let's focus on the interval $t\in [0, 1/2)$. Now you can ask that which matrices are in the form of $B_t$. You can solve for $B$ in terms of $B_t$ and that gives $B = ((1-t)B_t^T-tB_t)/(1-2t)$. This means that for any positive matrix $A$ and $t\in [0, 1/2)$ where $(1-t)A-tA^T>0$, we can find the appropriate $B$. So any positive matrix $A$ where we have $A>cA^T$ ($c$ here is $\frac{t}{1-t} \in [0, 1)$). So for any positive matrix $A$ where $A>cA^T$ such that $c\in [0, 1)$ we want to prove:

$$a^2 + (\frac{c+1}{c-1})^2b^2 \leq \rho(A)^2$$ Furthermore as $c$ get's larger the left side also gets larger so it suffices to prove the inequality for the maximum possible $c$ i.e we can just assume that $c = \min_{i, j}\frac{a_{i, j}}{a_{j, i}}$. This type of inequality is also reminded me of spectral gap problems I found a few papers on this that I found interesting and had completely new ideas to me (but they all deal with upper bounds for $\frac{|\lambda_2|}{|\lambda_1|}$). The first interesting paper is Cones and gauges in complex spaces: Spectral gaps and complex Perron-Frobenius theory. Take a look at Theorem 4.8 and the example following it. I am not too familiar with these ideas but they rely on the Hilbert metric on the cones. Another paper regarding an upper bounds for the spectral gap in terms of elements of the matrix can be found in A New Proof of Hopf’s Inequality Using a Complex Extension of the Hilbert Metric.