If $A,B\in GL_n(\mathbb{C})$ are positive matrices. Prove that the following holds

Question

If $A,B\in GL_n(\mathbb{C})$ are positive matrices. Prove that, there is $\lambda\in\sigma(A^{-1}B)$ such that the following holds $$\langle Ax,x\rangle\ge\lambda^{-1}\langle Bx,x\rangle\ \forall x\in\mathbb{C}^n$$

If A, B are simultaneously diagonalizable, then there exist orthonormal basis $\{u_1,\ldots,u_n\}$ of $\mathbb{C}^n$ and $p_i,q_i>0$ such that $A=\sum\limits_{i=1}^n p_i|u_i\rangle\langle u_i|$ and $B=\sum\limits_{i=1}^n q_i|u_i\rangle\langle u_i|$

Then I can choose $\lambda=\max\left\{\frac{q_i}{p_i}:\ i\in\{1,\ldots,n\}\right\}$. Obviously, $\lambda\in\sigma(W_1^{-1}W_2)$ and $A-\lambda^{-1}B=\sum\limits_{i=1}^n (p_i-\lambda^{-1}q_i)|u_i\rangle\langle u_i|$ and $p_i-\lambda^{-1}q_i\ge0$. Hence, we are done.

But I am not able to show it for general $A,B\in GL_n(\mathbb{C})$ positive. Can anyone help me in this regard? Thanks for your help in advance.

Answer 1

We have $$\sigma(A^{-1}B)=\sigma(B^{1/2}A^{-1}B^{1/2})= \sigma(B^{-1/2}AB^{-1/2})^{-1}$$ Let $\lambda$ be the largest element in $\sigma(A^{-1}B). $ Thus $\lambda^{-1}$ is the smallest element in $\sigma(B^{-1/2}AB^{-1/2}).$ Then $$\langle B^{-1/2}AB^{-1/2}y,y\rangle \ge \lambda^{-1}\langle y,y\rangle $$ Substituting $y=B^{1/2}x$ implies $$\langle Ax,x\rangle \ge \lambda^{-1}\langle Bx,x\rangle $$

Answer 2

Note that $A^{-1}B$ has the same spectrum as the similar matrix $A^{1/2}(A^{-1}B)A^{-1/2} = A^{-1/2}BA^{-1/2}$ (where $A^{1/2}$ denotes the unique positive "square root" of $A$). Note that $A^{-1/2}BA^{-1/2}$ is self-adjoint. We can further argue that $A^{-1/2}BA^{-1/2}$ is positive.

On the other hand, note that for a constant $k \in \Bbb R$, we have $$ \langle Ax, x \rangle \geq k\langle Bx,x \rangle \qquad \forall x \in \Bbb C^n \iff\\ \langle (A - kB)x,x\rangle \geq 0 \qquad \forall x \in \Bbb C^n \iff\\ \langle (A - kB)A^{-1/2}x,A^{-1/2}x\rangle \geq 0 \qquad \forall x \in \Bbb C^n \iff\\ \langle A^{-1/2}(A - kB)A^{-1/2}x,x\rangle \geq 0 \qquad \forall x \in \Bbb C^n \iff\\ \langle (I - kA^{-1/2}BA^{-1/2})x,x\rangle \geq 0\qquad \forall x \in \Bbb C^n. $$ In other words, $\langle Ax, x \rangle \geq k\langle Bx,x \rangle$ holding for all $x$ is equivalent to the statement that $I - kA^{-1/2}BA^{-1/2}$ is positive (but not necessarily invertible, i.e. positive semidefinite).

With that in mind, take $\lambda$ to be the largest element of $\sigma(A^{-1}B) = \sigma(A^{-1/2}BA^{-1/2})$. Argue that the spectrum of $\lambda^{-1}(A^{-1/2}BA^{-1/2})$ must be contained in the real interval $[0,1]$ (with its largest eigenvalue equal to $1$). It follows that the matrix $I - \lambda^{-1}(A^{-1/2}BA^{-1/2})$ has its spectrum within the interval $[0,1]$ (with its smallest eigenvalue equal to $0$).

Thus, the matrix $I - \lambda^{-1}(A^{-1/2}BA^{-1/2})$ is self-adjoint with a non-negative spectrum and is hence positive. Applying our earlier argument and taking $k = \lambda^{-1}$, we can conclude that $\langle Ax,x\rangle \geq \lambda^{-1}\langle Bx,x \rangle$ holds for all $x \in \Bbb C^n$, which was what we wanted.

Answer 3

Let $\langle x,y \rangle_B := \langle Bx,y \rangle$. Then $B^{-1}A$ is selfadjoint w.r.t. $\langle \cdot,\cdot \rangle_B$. By Rayleigh-Ritz $$\frac{1}{\lambda}:=\inf \sigma(B^{-1}A) = \inf_{x\in \mathbb{C}^n\smallsetminus\{0\}} \frac{\langle B^{-1}Ax,x\rangle_B}{\langle x, x \rangle_B}$$ Therefore $\lambda \in \sigma(A^{-1}B)$ and $\frac{1}{\lambda}\langle Bx,x \rangle = \frac{1}{\lambda}\langle x,x \rangle_B \leq \langle B^{-1}Ax,x \rangle_B = \langle Ax,x \rangle$ for all $x \in \mathbb{C}^n$.