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:

1. if the inverse of a positive definite matrix is positive definite

2. if the matrix square root is positive definite, or symmetric

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

4. whether the product is symmetric

Can someone clarify this for me using some well known facts about symmetric matrices?

• how do you calculate $A^{-1/2}$ for such a matrix? – janmarqz Nov 23 '15 at 3:43
• A simmetric positive-defined matrix does not have a unique square root. It has a unique symmetric and positive-definite square root. – user228113 Nov 23 '15 at 4:00

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.

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.