$A+B+\frac{1}{2}BA^{-1}B$ is positive definite.

Question

Given symmetric positive definite matrix $A \in \mathbb{R}^{n\times n}$ and symmetric matrix $B\in \mathbb{R}^{n\times n}$, how to prove that $X = A+B+\frac{1}{2}BA^{-1}B$ is positive definite?

This can be easily seen when they are both diagonal: the diagonal entries of $X$ is

$$x_{ii} = a_{ii}+b_{ii}+\frac{b_{ii}^2}{2a_{ii}} > 0$$ for all $b_{ii} \in \mathbb{R}$ and $a_{ii} > 0$. But I am struggling a bit proving the general case.

Answer 1

The proof follows similar pattern that the diagonal case. For example how did you prove it for the diagonal case? One way of doing it is rewriting $x_{ii}$ as:

\begin{align} x_{ii} = \frac12 a_{ii}^{\frac12}\left(1+\left(1+\frac{b_{ii}}{a_{ii}}\right)^2\right)a_{ii}^{\frac12} \end{align}

In the general case you can rewrite $X$ as:

$$X = \frac12 A^{\frac12} \left(I + \left(I + A^{-\frac12}BA^{-\frac12}\right)^2\right)A^{\frac12}$$

Can you complete the proof from here?

Answer 2

$$A+B+\frac{1}{2}B{A^{-1}}B=\frac{1}{2}A+\frac{1}{2}A^{1/2}(I+A^{-1/2}BA^{-1/2})^2A^{1/2}\geq 0.$$