Show derivative of matrix map is surjective

Question

Suppose I have the map $h\colon M_n(\mathbb{R})\to S_n(\mathbb{R})$, $X\mapsto XX^T$, how would I compute its derivative to show that it's surjective? (Here, $S_n(\mathbb{R})$ are the symmetric matrices.)

Answer 1

$h(X+H)=(X+H)(X+H)^T=(X+H)(X^T+H^T)=XX^T+XH^T+HX^T+HH^T.$

It follows that the derivative is given by $h'_X=X(\cdot)^T+(\cdot) X^T$.

It is not surjective is general (for instance, $h'_0=0$). Don't you want to restrict to some particular subset to compute the derivative?

If you want to show $I$ is a regular value, let $X$ be in $h^{-1}(I)$. Now, let $A$ be a symmetric matrix. Verify that $h'_X(\frac{1}{2}AX)=A.$ This shows surjectivity.