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Let $A, B$ be symmetric positive definite matrices. Suppose that $\frac{1}{2}B^{1/2} \preceq A \preceq B^{1/2}$. Can we prove $AB^{-1}A\preceq \text{constant}\cdot I\,\,$?

If we could have $A^2 \preceq B$ from $A \preceq B^{1/2}$, then we have $B^{-1}\preceq (A^2)^{-1} = A^{-2},$ and can show $AB^{-1}A\preceq A(A^{-2})A = I$.

Unfortunately, by Is $A^2-B^2$ positive definite too when $A-B,B$ is positive definite?, $A \preceq B^{1/2}$ does not imply $A^2 \preceq B$.

I am thinking that the other inequality $ A \succeq \frac{1}{2}B^{1/2}$ might help. Also, I can allow the "constant" to be some larger constant than $1$. However, it is still unclear how to show the desired inequality.

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Bob Wei
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1 Answers1

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With some numerical experiment, I found this is false, though there is no formal proof. For $n\geq 1$, set
$B^{\frac{1}{2}}=\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix}$ and generate $A$ according to the following: $A=\frac{1}{2}B^{\frac{1}{2}} + A'$ where $A'=r\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$ with scalar $r>0$ set to the largest number that makes $A'\preceq \frac{1}{2}B^{\frac{1}{2}}$, and $\theta$ set to a fixed number (e.g., $\theta=\frac{\pi}{256}$).

This clearly satisfies the specified condition. But experiment will show that the largest eigenvalue of $AB^{-1}A$ scales (linearly) with $n$.

Bob Wei
  • 21