Rank of a Differential This a mostly a sanity check sort of thing. I am working with the map given by 
$$F:M_{2\times 2}(\mathbb{R})\rightarrow S_{2\times 2}(\mathbb{R}):A\mapsto A^t J A$$
where J is the matrix
$$J = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}.$$
I am trying to show that the set $S = \{A : F(A) = J\}$ is a smooth submanifold of $M$. I know that I need to use the Regular Level Set Theorem to do this, but I am having trouble showing $DF$ will have maximal rank for that the matrices in $S$ (which I have already shown contains only invertible matrices). Using the canonical isomorphism on $M_{2\times 2}(\mathbb{R})\cong \mathbb{R}^4$ described by $\begin{pmatrix}a_1 & a_2\\ a_3 & a_4\end{pmatrix}\mapsto (a_1,a_2,a_3,a_4)$ and the similar one on $S_{2\times 2}(\mathbb{R})\cong \mathbb{R}^3$, I computed
$$F(a_1,a_2,a_3,a_4) = (a_1^2 - a_3^2, a_1a_2 - a_3a_4, a_2^2 - a_4^2)$$
which gives us that
$$DF = \begin{pmatrix} 2a_1 & 0 & -2a_3 & 0\\
                       a_1 & a_2 & -a_3 & -a_4\\
                       0 & 2a_2 & 0 & -2a_4\end{pmatrix}.$$
My problem is that I think that this matrix can never be full rank since the center row can always be eliminated. Have I done something wrong here? Can anyone verify for me under what conditions $DF$ would be surjective?
 A: Slightly more abstractly (rather than resorting to coordinates): 
Note that $df_A(B) = B^\top JA + A^\top JB$. Given a arbitrary symmetric matrix $C$, we wish to find $B$ so that $df_A(B) = C$. Since you already observed that any $A\in f^{-1}(J)$ must be invertible, take
$$B = \tfrac12 J(A^{-1})^\top C,$$
and check that $df_A(B) = C$, as desired.
The advantage of this sort of approach is that it will work nicely in higher dimensions, whereas the coordinate approach is very awkward.
A: Ok, I just wanted to add a post showing another way of answering this question that is more in line with the structure posted in the question itself. This is NOT an elegant solution like the one provided above by Ted Shifrin (Thank you again!), but I think it might be useful for some people seeing this later to have both methods available to them.
Anyway, the matrix for $DF$ that is provided above is wrong. The correct matrix for the differential is given by
$$DF = \begin{pmatrix}
2a_1 & 0 & -2a_3 & 0\\
a_2 & a_1 & -a_4 & -a_3\\
0 & 2a_2 & 0 & -2a_4
\end{pmatrix},$$
and to use the Regular Level Set Theorem, we need the map $DF$ to be surjective which is equivalent to the matrix representation of $DF$ having maximal rank (3 in this case). There are two ways for the above matrix to drop rank: either we eliminate the middle row, or we eliminate the right two columns. Using row operations we can see that an attempt to reduce the middle row would result in the matrix 
$$\begin{pmatrix}
2a_1 & 0 & -2a_3 & 0\\
2(a_1 + a_2) & 2(a_1 + a_2) & -2(a_3 + a_4) & -2(a_3 + a_4)\\
0 & 2a_2 & 0 & -2a_4
\end{pmatrix}$$
which will drop rank precicely when $a_1 = -a_2$ and $a_3 = -a_4$ $(*)$. Similarly, an attempt to reduce the right two columns will result in the matrix
$$\begin{pmatrix}
2a_1 & 0 & 2(a_1 - a_3) & 0\\
a_2 & a_1 & a_2-a_4 & a_1-a_3\\
0 & 2a_2 & 0 & 2(a_2 - a_4)
\end{pmatrix}$$
which will drop rank when $a_1 = a_3$ and $a_2 = a_4$ $(**)$. If a matrix $A$ satisfies $(*)$ or $(**)$, we will have that $\det(A) = 0$, so, as long as all of the matrices $A$ that satisfy the condition $A^tJA = J$ do not have determinant 0, we will have that $J$ is a regular value and the set $S = \{A : A^tJA = J\} = F^{-1}(J)$ is a regular level set. This does turn out to be the case, as can be verified using properties of the determinant, so $S$ is a smooth submanifold by the Regular Level Set Theorem as desired. 
