If $A$ and $B$ are real, symmetric, positive definite matrices, then prove that $A(A+B)^{-1}B$ is positive definite

Question

If $A$ and $B$ are real, symmetric, positive definite matrices, then prove that $A(A+B)^{-1}B$ is positive definite.

If $A$ and $B$ are positive definite, then $A+B$ is positive definite, and inverse of a positive definite matrix is positive definite. So $(A+B)^{-1}$ is positive definite. I don't know how to proceed after that.

Please anyone help me solve it. Thanks in advance.

Answer 1

It's true because $A(A+B)^{-1}B=\left[B^{-1}(A+B)A^{-1}\right]^{-1}=(B^{-1}+A^{-1})^{-1}$ and the sum or inverses of any symmetric positive definite matrices are symmetric positive definite.

Answer 2

Note that $M = A(A + B)^{-1}B$ is positive definite if and only if $M + M^T$ is symmetric and positive definite. Also, $(A + B)^{-1}$ is symmetric since it is the inverse of a symmetric matrix. Now,

Note that $$ A(A + B)^{-1}B = \\ (A+B-B)(A+B)^{-1}B = \\ (A+B)(A+B)^{-1}B - B(A+B)^{-1}B = \\ B - B(A + B)^{-1}B = \\ B[B^{-1} - (A + B)^{-1}]B $$ Now, it suffices to show that $B^{-1} - (A + B)^{-1}$ is positive semidefinite. This proof is quick if you're familiar with the Loewner ordering, but one direct proof would be as follows. Note that $$ A \quad \text{is positive semidefinite} \iff\\ (A + B) - B \quad \text{is positive semidefinite} \iff\\ B^{-1/2}(A + B)B^{-1/2} - I \quad \text{is positive semidefinite} \iff\\ B^{-1/2}(A + B)B^{-1/2} \quad \text{is symmetric with eigenvalues greater than $1$} \iff\\ [B^{-1/2}(A + B)B^{-1/2}]^{-1} \quad \text{is symmetric with eigenvalues less than $1$} \iff\\ B^{1/2}(A + B)^{-1}B^{1/2} \quad \text{is symmetric with eigenvalues less than $1$} \iff\\ I - B^{1/2}(A + B)^{-1}B^{1/2}\quad \text{is positive semidefinite} \iff\\ B^{-1} - (A + B)^{-1} \quad \text{is positive semidefinite} $$

Answer 3

Given $\rm A, B \succ \mathrm O_n$, we form the following block matrix

$$\begin{bmatrix} \mathrm A + \mathrm B & \mathrm A\\ \mathrm A & \mathrm A\end{bmatrix} = \begin{bmatrix} \mathrm I_n & \mathrm I_n\\ & \mathrm I_n\end{bmatrix} \underbrace{\begin{bmatrix} \mathrm B & \\ & \mathrm A\end{bmatrix}}_{\succ \mathrm O_{2n}} \begin{bmatrix} \mathrm I_n & \\ \mathrm I_n & \mathrm I_n\end{bmatrix} \succ \mathrm O_{2n}$$

and, using the Schur complement, we conclude that

$$\mathrm A - \mathrm A \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A \succ \mathrm O_n$$

Note that

$$\mathrm I_n = \left( \mathrm A + \mathrm B \right)^{-1} \left( \mathrm A + \mathrm B \right) = \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A + \left( \mathrm A + \mathrm B \right)^{-1} \mathrm B$$

and, hence,

$$\left( \mathrm A + \mathrm B \right)^{-1} \mathrm B = \mathrm I_n - \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A$$

Left-multiplying both sides by $\rm A$,

$$\mathrm A \, \left( \mathrm A + \mathrm B \right)^{-1} \mathrm B = \mathrm A - \mathrm A \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A \succ \mathrm O_n$$