Showing that positive definite symmetric matrices have a smooth square root

Question

Show that a map in the vector space of positive-definite symmetric $n \times n$ matrices
$$f:S_n^+(\mathbb{R}) \rightarrow S_n^+(\mathbb{R})$$ $$f(s) = s^2$$ is a smooth diffeomorphism.

My thought is that if I can show that the derivative $f'(a)(b)$ is invertible for all $s \in S_n^+(\mathbb{R})$, then I will have shown that there is a positive definite square root that varies smoothly with $s$.

Since $s$ is a symmetric positive definite matrix, we know that it has $n$ eigenvectors with positive eigenvalues. Further, we know that $$f'(a)(b) = ab + ba$$

$$ ab + ba = 0 $$ $$ abb^T + bab^T = 0 $$

Suppose that $b$ has non-zero entries. We know that $a, bb^T$ are both positive definite matrices, and we know that diagonal entries of $bab^T$ are positive, since they are equal to $b_iab_i^T$, which is positive because $a$ is positive definite. How do I show that the product $abb^T$ has non-negative entries on the diagonal? I want to show that this results in a contradiction.

Answer 1

The last bit, that needs to be proven is: Let $A$ be symmetric and positive definite. Let $B$ be a matrix such that $AB + BA = 0$. Then it follows that $B=0$.

Since $A$ is symmetric and positive definite, there exists a matrix $Q$ with $QAQ^{-1} = D = {\rm diag} (d_1,\dotsc,d_n)$ with $d_i>0$ for all $i$. Putting $C = (c_{i,j})_{i,j}:= QBQ^{-1}$, we have \begin{align*} 0 &= QAQ^{-1} QBQ^{-1} + QBQ^{-1} QAQ^{-1} = DC + CD \end{align*} or equivalently, $DCD^{-1} = -C$. But now we have $-c_{i,j} = d_id_j^{-1} c_{ij}$ and hence $c_{i,j}=0$ because of $d_id_j^{-1}>0$. This shows $C=0$ and therefore $B=0$.

Answer 2

Some points:

First you say you are working in the vector space of (real) positive definite symmetric matrices, this is not a vector space (as you correctly note when you say $0\notin S_n^+$). This point is largely irrelevant.

You are looking at $f$ as a map $f: S_n^+\to S_n^+$, here the derivative $f'(a)$ is a map between the tangent spaces $f'(a): T_aS_n^+\to T_aS_n^+$. It is right that you want this map to be invertible, but you cannot identify $T_a S_n^+$ with $S_n^+$. Instead you must calculate that $T_aS_n^+$ can be identified with the vector space of symmetric matrices.

To see it let $\alpha>0$ so that $x^Ta\, x ≥ \alpha x^2$ for all $x\in\mathbb R^n$, now note: $$x^T (a+\epsilon b) x = x^T a\, x+\epsilon\, x^T b \, x ≥ \alpha\, x^2-\epsilon\, \|b\|\,x^2$$

So for any symmetric matrix $b$ there is an $\epsilon>0$ so that $a+\epsilon b\in S_n^+$, this will give you a canonical inclusion of the space of symmetric matrices into $T_aS_n^+$. However the space of symmetric matrices is obviously of higher or equal dimension as $T_aS_n^+$ and thus the inclusion must be an isomorphism.

For the rest your current proof doesn't quite work, but it is the statement that the only skew symmetric matrix that is a product of a positive definite and a symmetric matrix must be $0$. I believe that is correct but I'm thinking of a proof.