# If A,B are real symmetric positive definite matrices, then $B^{-1} AB$ is symmetric positive definite.

If $A,B$ are real symmetric positive definite matrices, then $B^{-1} AB$ is symmetric positive definite.

I know that, generally, product of symmetric positive definite matrices need not be even symmetric. But what if we premultiply it by the inverse of one of matrices?

• no, $(B^{-1}AB)^T = B^T A^T (B^{-1})^T = BAB^{-1}$, which is not equal to $B^{-1}AB$ unless you happen to have $B^{-1} = B$ – cats May 13 '15 at 5:46
• It holds when $B$ is an orthogonal matrix – Alonso Delfín May 13 '15 at 5:51
• @AlonsoDelfín A symmetric positive definite orthogonal matrix $B$ is the identity. – egreg May 13 '15 at 7:21
• @egreg you are right, and of course it holds when B is de identity. – Alonso Delfín May 13 '15 at 12:38

Note that $B^{-1}$ is also SPD and we have that $$(B^{-1}AB)^\top = B^\top A^\top (B^{-1})^\top = BAB^{-1}$$ which shows that $B^{-1}AB$ is not necessarily symmetric in general. In particular, consider \begin{align} A &= \begin{pmatrix}1 & 1\\1 & 3\end{pmatrix}\\ B &= \begin{pmatrix}2 & 1\\1 & 3\end{pmatrix}\\ \end{align} Then \begin{align} B^{-1}AB &= \begin{pmatrix}0.8 & 0.4\\1.4 & 3.2\end{pmatrix}\\ \end{align} is not symmetric.
If $A$ is a $m \times m$ real symmetric positive definite matrix and $B$ is a $m \times n$ real matrix, then $B^\top AB$ is a $n \times n$ real symmetric positive semidefinite matrix. In particular $B^\top AB$ is real symmetric positive definite if and only if $B$ is full column rank.
The proof is quite straightforward: $B^\top AB$ is symmetric since $$(B^{\top}AB)^\top = B^\top A^\top (B^{\top})^\top = B^\top AB$$ and $B^\top AB$ is positive semidefinite since for $x \in \mathbb{R}^n \setminus \{0\}$, $$x^\top(B^\top AB)x = (Bx)^\top A (Bx)^\top \geq 0$$ The inequality can be made strict if and only if $Bx \neq 0$ for all $x \neq 0$, which means $N(B) = \{0\}$ and $rank(B) = n$.