What are the regular values of the map $A\mapsto A^{t}A$ Let $M(n,\mathbb{R})$ denote the set of real $n\times n$ matrices and $\text{Sym}(n,R)$ denote the set of real $n\times n$ symmetric matrices. Consider theses set as $\mathbb{R}^{n^{2}}$ and $\mathbb{R}^{n(n+1)/2}$, respectively. The problem is this: find all regular/critical points and regular/critical values of the map $f(A)=A^{t}A$. I found that $d_{A}f(B)=A^{t}B+B^{t}A$. But I can't determine when $d_{A}f$ is surjective. I also found that $d_{A}f$ is surjective if $A$ is nonsingular. But I'm not sure this is a necessary condition for $d_{A}f$ to be surjective. How can I find all matrices so that $d_{A}f$ to be surjective?
 A: Write $f(A) = A^TA$. 
Note that $A^TA$ is always a symmetric non-negative matrix, so there is an orthogonal matrix $O$ so that $A^{T}A = O^T DO$, where $D$ is a diagonal matrix with non-negative diagonal entries. Let $C = O^T \sqrt D O$. Then $f(C) = O^T D O = f(A)$. Consider
$$d_Cf (B) = C^T B + B^T C.$$
If we assume that $A$ is singular, then $D$ has a zero diagonal entries (at $(j,j)$ for example). Then $d_Cf$ is not surjective: Let $J$ be the matrix with a nonzero entry at $(j,j)$. Then  
$$\begin{array}{crcl} & C^T B + B^T C &=& O^T JO \\
\Leftrightarrow & (O^T\sqrt D O)^T B + B^T O^T \sqrt DO &=& O^T JO \\
\Leftrightarrow & (O^T\sqrt D) (O B) + (OB)^T \sqrt DO &=& O^T JO \\
\Leftrightarrow & \sqrt D (O B) + O(OB)^T \sqrt DO &=&  JO \\
\Leftrightarrow & \sqrt D (O B O^T)+ (OBO^T)^T \sqrt D &=&  J 
\end{array}$$
But this is impossible as the $j$-row and $j$-th column of $\sqrt D$ are zero (so the $(j, j)$-entry of the left hand side is zero). Hence $d_Cf$ is not surjective. 
Thus by the definition of a regular value, $f(A)$ is not a regular value if $A$ is non-singular. (Note that I haven't checked if $d_Af$ is surjective. We don't need to check that)
A: Proposition. The non-singular matrices are regular points of $f$. 
Proof. Let $S_n$ be the set of symmetric matrices and  $S_n^+$ be the set of symmetric positive matrices which is open in $S_n$. Note that $f_{|GL_n}: GL_n(\mathbb{R})\rightarrow S_n^+$ is onto and $C^{\infty}$.
Let $g:V\in S_n^+\rightarrow L\in T_n$ where $V=L^TL$ is the Cholesky decomposition of $V$ and $T_n$ is the set of upper triangular matrices with positive diagonal. Then $g$ is a $C^{\infty}$ diffeomorphism and is a right inverse of $f_{|GL_n}$: $f_{|GL_n}\circ g=id_{S_n^+}$. Consequently $D(f_{|GL_n})\circ Dg=id$ and  $f_{|GL_n}$ has no critical points on $T_n$.
Now let $A\in GL_n$ and $L=g\circ f(A)$. There is $O\in O_n$ s.t. $A=OL$ (cf. QR decomposition). Thus $Df_A(H)=L^TO^TH+H^TOL=L^TK+K^TL=Df_L(K)$ where $H\rightarrow K=O^TH$ is a linear isomorphism. Finally $rank(Df_A)=rank(Df_L)=n(n+1)/2$ and we are done.
