Consider the map $f: \mathbb{R}^{n^2} \rightarrow \mathbb{R}^{\frac{n(n+1)}{2}}$, i.e. a map from $n\times n$ matrices with real entries to $n\times n$ symmetric matrices, defined by $f(X)=X^tX$.
I know that the differential map of $f$ is given by $Df(X)Y=X^tY+Y^tX$. I am trying to show that for any invertible matrix $X\in M(n, \mathbb{R})$, $$\dim(\ker(Df(X)))=\frac{n(n-1)}{2}$$
I know that we must show that $Df(X)$ is an onto/surjective map, so for any symmetric matrix $S$, I can find $T$ such that $[Df(X)]T=X^tT+T^tX=S$, but I am not sure how to use the invertibility of $X$ to show this.
Let $Y\in \ker Df(X)$ and $A = X^tY$. Note that $X^tY+Y^tX = A+A^t = 0$ thus $A$ is antisymmetric. If you know the dimension of the space of antisymmetric matrices, then it's almost OK. Note also that $Y\mapsto A = X^tY$ is a linear bijection thus $\ker Df(X) \simeq \{A \colon A^t = -A\}$.
Edit: (to answer the comment below)
If you prefer, you can show that $Df(X)$ is onto. Using the same trick $A=X^tY$, you could show that $Y\mapsto X^tY+Y^tX = A+A^t$ is surjective from the $n\times n$ matrices to the $n\times n$ symmetric matrices. If $S$ is a symmetric matrix and if you take $Y$ such that $A=\dfrac{S}{2}$, i.e. $Y = (X^t)^{-1}\dfrac{S}{2}$, then this $Y$ satisfies $Df(X)Y = \dfrac{S}{2} + \dfrac{S}{2} = S$.