Estimate parameter $a$ such that $tr \left[ A (B- (I-aC)B(I-aC) ) \right] > 0$.

Question

Suppose $A, B, C$ are all real symmetric and positive definite matrices. Consider the function $f: \mathbb R \to \mathbb R$ given by $$ a \mapsto {\bf tr}\left[ A (B- (I-aC)B(I-aC) ) \right],$$ where $I$ is identity matrix. It is clear $f(0) = 0$ and further assume there exists some $\tau > 0$ such that $f(x) > 0$ for every $x \in (0, \tau)$. We may as well assume the maximal interval such that $f(x) > 0$ to be $(0, \tau)$. That is, $f(x) > 0$ for $x \in (0, \tau)$ and $f(0) = f(\tau) = 0$. I am wondering with these information, is it possible to deduce $\tau \ge \frac{1}{\lambda_{\max}(C)}$?

Essentially I am in the situation that I know for small $a$, the trace is positive and by continuity there should be some maximal interval the trace is always positive. I want to estimate this interval. I tried to use a crude bound \begin{align*} {\bf tr}\left[ A (B- (I-aC)B(I-aC) ) \right] \ge \lambda_{\min}(A) {\bf tr}(B) - \lambda_{\max}^2(I-aC)\lambda_{\max}(A) {\bf tr}(B). \end{align*} But this gives us meaningless bound since if we set above bound to be greater tha $0$, $a$ could be possibly unsolvable.

On the other hand, I feel that $a$ must be related to $\lambda_{\max}(I-aC)$ so we can choose $a$ to minimize this quantity and this would give us $a'=\frac{2}{\lambda_{\min}(C) + \lambda_{\max}(C)}$. Intuitively, I would imagine over $[0, a']$, $f$ should be positive.

Answer 1

$\DeclareMathOperator{\Tr}{Tr}$ Let $$\begin{align} D\stackrel{\text{def}}{=}\sqrt{C}A\sqrt{C}\\ E\stackrel{\text{def}}{=}\sqrt{C}B\sqrt{C}\\ F\stackrel{\text{def}}{=}C^{-1}\\ H\stackrel{\text{def}}{=}DE+ED \end{align}$$ Then

• $D$, $E$, and $F$ are positive-definite
• $H$ is Hermitian, and
• if $\tau$ exists, it is determined by $$\frac{\tau}{2}=\frac{\Tr HF}{\Tr H}\text{.}$$

Now, if we somehow knew that $H$ were positive-definite whenever $D$ and $E$ are, then we would be able to conclude $$\frac{\Tr HF}{\Tr H}\stackrel{?}{\geq}\lambda_{\text{min}}(F)\text{.}$$

But $H$ need not be positive definite: consider $D=\tfrac{1}{2}(1+r\sigma_x)$, $E=\tfrac{1}{2}(1+r\sigma_z)$ where $\tfrac{1}{2}<r^2<1$ and $\sigma_x,\sigma_z$ are Pauli matrices.

Instead, let $H=H^+-H^-$ be the Hahn–Jordan decomposition of $H$. ($H^+$ is the sum of over the positive-eigenvalue terms in the spectral decomposition of $H$, and $H^-$ is the negative of the sum of the negative-eigenvalue terms; $H^+$ and $H^-$ are both semidefinite.) Assume without loss of generality that $\Tr H>0$.

Then $$\begin{split}\frac{\Tr HF}{\Tr H}&=\frac{\Tr (H^+-H^-)F}{\Tr H}\\ &=\frac{\Tr H^+F}{\Tr H}-\frac{\Tr H^-F}{\Tr H}\\ &=\frac{\Tr (H^+\lambda_{\text{min}}(F)-H^-\lambda_{\text{max}}(F))}{\Tr H}+\frac{\Tr H^+(F-\lambda_{\text{min}}(F))}{\Tr H}+\frac{\Tr H^-(\lambda_{\text{max}}(F)-F)}{\Tr H}\\ &\geq\frac{\Tr (H^+\lambda_{\text{min}}(F)-H^-\lambda_{\text{max}}(F))}{\Tr H}\\ &=\lambda_{\text{min}}(F)-(\lambda_{\text{max}}(F)-\lambda_{\text{min}}(F))\frac{\Tr H^-}{\Tr H}\text{.} \end{split}$$ Note that this bound is weaker than the one we would get if $H$ were known to be positive-definite. It is actually achieved for $$F=\lambda_{\text{min}}(F)[H>0]+\lambda_{\text{max}}(F)[H <0]$$ where $[H>0]$ and $[H<0]$ are the projectors onto the positive and negative part of the spectrum of $H$.