Is the following product of matrices symmetric? Suppose I have $A,B$, $A$ and $B$ both positive definite and symmetric
Then is the following product symmetric?
$$A^{-1/2}BA^{-1/2}$$
I understand that $A$ is always invertible, and has a unique square root.
I don't know:


*

*if the inverse of a positive definite matrix is positive definite

*if the matrix square root is positive definite, or symmetric

*if the inverse of the matrix square root is positive definite or symmetric

*whether the product is symmetric
Can someone clarify this for me using some well known facts about symmetric matrices?
 A: It is not true that a positive definite matrix has a unique square root.  What is true is that it has a unique square root that is also positive definite (that is, it has many square roots, but only one of its square roots is positive definite).
When someone writes $A^{1/2}$ for positive definite $A$, they usually mean the unique positive definite square root of $A$ (much as when you write $\sqrt{x}$ and $x$ is a positive number, you usually mean the positive square root of $x$).  So from now on, I will assume that is what $A^{1/2}$ means in your question.  If $A^{1/2}$ is just any square root of $A$, then there is no reason for $A^{-1/2}BA^{-1/2}$ to be positive definite.
Now let's observe that the inverse of a positive definite matrix $C$ is positive definite.  First, since $(CD)^T=D^TC^T$, letting $D=C^{-1}$, we see that $(C^{-1})^T=(C^T)^{-1}$.  If $C$ is symmetric, it follows that $C^{-1}$ is symmetric as well.  Now $C$ is positive definite iff it is symmetric and has positive eigenvalues, but the eigenvalues of $C^{-1}$ are just the inverses of the eigenvalues of $C$ (with the same eigenvectors).  So $C$ has positive eigenvalues iff $C^{-1}$ has positive eigenvalues.  We conclude that if $C$ is positive definite, so is $C^{-1}$.
So from this, we see that $A^{-1/2}$ is positive definite.  To show that $A^{-1/2}BA^{-1/2}$ is positive definite, let us consider the unique positive definite square root $B^{1/2}$ of $B$.  Notice that we can write $A^{-1/2}BA^{-1/2}=(A^{-1/2}B^{1/2})(B^{1/2}A^{-1/2})$.  Writing $C=A^{-1/2}B^{1/2}$, note that $C^T=(B^{1/2})^T(A^{-1/2})^T=B^{1/2}A^{-1/2}$, since $B^{1/2}$ and $A^{-1/2}$ are symmetric.  Thus we have $A^{-1/2}BA^{-1/2}=C^TC$.  But for any invertible matrix $C$, $C^TC$ is positive definite, and our $C$ is invertible since $A^{-1/2}$ and $B^{1/2}$ are.  Thus $A^{-1/2}BA^{-1/2}$ is positive definite.
A: The inverse of a positive definite matrix is positive definite. A matrix is positive definite if it has positive eigenvalues and the eigenvalues of $A^{-1}$ are just the inverse of the eigenvalues of $A$ (see e.g. this answer).

If $A$ is positive definite and symmetric then $A^{1/2}$ does not have to be either symmetric or positive definite. Example: $A^{1/2} = \pmatrix{0 & 2\\ \frac{1}{2} & 0} \implies A = \pmatrix{1 & 0 \\0 & 1}$.

For the product $A^{-1/2}BA^{-1/2}$ let $A=I$ and $B = \pmatrix{2a & a\\a & 2a}$. If $a>0$ then both matrices are positive definite and symmetric. A non-symmetric square root of $A$ is $A^{-1/2}=\pmatrix{0 & 2\\ \frac{1}{2} & 0}$ for which $A^{-1/2}BA^{-1/2} = \pmatrix{2a & 4a\\\frac{a}{4} & 2a}$ which is not symmetric.
