Tangent Space of SymSL(n,$\mathbb{R}$) at arbitrary point I am looking for the tangent space of $SymSL(n,\mathbb{R}) = \{A\in\mathbb{R}^{n\times n} | \,A^{\rm T} = A,\; \det{A}=1 \}$ (actually $n=3$) at an arbitrary $M\in SymSL(n,\mathbb{R})$.
(Actually I am interested in $SymSL^+(n,\mathbb{R}) = \{A\in\mathbb{R}^{n\times n} | \,A^{\rm T} = A,\; \det{A}=1,\; A \text{ is positive definite} \}$, but I have the feeling it boils down to the title of this question.)

Related questions:
Tangent Space of SL(n,ℝ) at arbitrary point, e.g. not at $\mathbb{1}$
Tangent space of Sym(n,ℝ)

The first question's answer doesn't work here because $Sym(n)$ is not a group, the second question's answer is not applicable because $SL(n)$ is not a vector space. Hence $SymSL(n,\mathbb{R})$ is neither a group nor a vector space.
I would very much appreciate any comments.
edit #2: $SL(n,\mathbb{R})$ is connected and smooth. So $SymSL(n,\mathbb{R})$ is the intersection of a connected smooth manifold with a vector space. This makes me believe we can talk about a tangent space in this case. Some more possibly usefull facts: the set $SymSL^+(n,\mathbb{R})$ I am actually interested in is a connected, simply connected and complete Riemannian manifold.
 A: Let $\mathcal{S}$ be the vector space (of dimension $k=n(n+1)/2$) of the real symmetric matrices. We consider the algebraic set $V=\{X\in\mathcal{S}| \det(X)=1\}$. Let $A\in V$. Then the tangent space of $V$ in $A$ is $T_A(V)=\{H\in \mathcal{S}|trace(HA^{-1})=0\}$.
Proposition. $V$ has dimension $k-1$.
Proof. This is equivalent to show that $dim(T_A(V))=k-1$ or that the linear form $f:H\in\mathcal{S}\rightarrow trace(HA^{-1})$ is not identically zero; this last point is clear because $f(A^{-1})=trace(A^{-2})>0$.
Remark. Note that $W=\{X\in V|X>0\}$ is open in $V$. Then, the tangent space of $W$ in $A\in W$ is the same as the tangent space of $V$ in $A$. Of course, $dim(W)=k-1$.
EDIT. Answer to @ Dr Doolittle . Let $k\leq n$, $f:\mathbb{R}^n\rightarrow \mathbb{R}^k\in C^1$ , $V=\{x\in\mathbb{R}^n|f(x)=0\}$ and $a\in V$. If $Df_a$ has full rank, then, in a neighborhood of $a$, $V$ is a variety of dimension $n-k$ and the tangent space of $V$ in $a$ is $T_a(V)=\{h\in\mathbb{R}^n|Df_a(h)=0\}$. 
Here $f(X)=\det(X)-1$ and $Df_A(H)=trace(HA^{-1})$.
A: The following is a rather trivial fact that seems to immediately provide a partial solution. Furthermore, this partial solution may be in fact a complete one in many cases.
Let $N$ be a smooth manifold, and let $L,M\subset N$ be submanifolds. Let $p\in L\cap M$ such that around $p$ the intersection $L\cap M$ is smooth. Then$$T_p(L\cap M)\subset \left(T_pL\right)\cap\left(T_pM\right).$$If $L$ and $M$ intersect transversally, the above $\subset$ is an equality.
Edit: A bit more on transversality: We say $L$ and $M$ intersect each other transversally if for every $p\in L\cap M$ we have$$T_pN=T_pL+T_pM.$$The most important fact about transversality is that whenever $L$ and $M$ intersect each other transversally, it automatically follows that $L\cap M$ is a smooth submanifold. Additionally, we also have $$T_p(L\cap M)=(T_pL)\cap(T_pM)$$for any $p\in L\cap M$. In particular, we know the dimension of $L\cap M$.
A: I am interested in seeing an answer to this question, I haven't been able to get an answer but this is what I do have to help prompt a conclusive answer. 
First as noted in the OP the space of positive symmetric matrices is a Riemannian manifold with tangent space at a particular point $\mathbf{P}$ given by $\{ \mathbf{P} \} \times \mbox{Sym}(n,\mathbb{R})$. We are interested in the variety defined by $\{ \mathbf{P} \; : \; \mbox{det}(\mathbf{P}) = 1 \}$, using the point in Amitai's answer leads to
$$
T_{\mathbf{P}} \mathcal{M} \subset \mbox{Sym}(n,\mathbf{R}) \cap \{ \mathbf{M} \; : \; \mbox{tr}(M) = 0\}.
$$ 
Proving one way or another that subset is proper or not I haven't been able to do, but here is a motivating example in $\mathbb{R}^{2\times 2}$, I also define the symmetric, trace zero matrices
$$
\mathbf{E}_1 = \begin{bmatrix} 1 & 0\\0 &-1 \end{bmatrix}, \qquad \mathbf{E}_2 =  \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.
$$
Now we want 
$$
 0 = \frac{d}{dt}\mbox{det}\,\mathbf{A}(t) = \mbox{tr}\left(\mbox{adj}\, \mathbf{A}(t) \frac{d A}{dt} \right),
$$
or
$$
\mbox{adj} \, A(t) \frac{dA}{dt} = c_1 \mathbf{E}_1 + c_2\mathbf{E}_2,
$$
for some constants $c_1$ and $c_2$, although I guess $c_1, c_2$ could be time dependent also? Anyway working with this I get
\begin{align}
 \frac{dA}{dt} = \frac{1}{\mbox{det}(\mbox{A}) } \mbox{A} \left(c_1\mathbf{E}_1 + c_2 \mathbf{E}_2 \right).
\end{align}
In particular we should have $\mathbf{P} = \mathbf{A}(0)$ and therefore we should be able to find constants $c_1$ and $c_2$ such that
$$
\mathbf{P} \left(c_1 \mathbf{E}_1 + c_2 \mathbf{E}_2 \right) = \mathbf{E}_i, \qquad i = 1,2
$$
a particular instance of this when $i=1$ implies
$$
\begin{bmatrix}
c_1 & c_2 \\
c_2 & -c_1 
\end{bmatrix} = 
\begin{bmatrix}
p^{-1}_{11} & -p^{-1}_{12} \\
p^{-1}_{12} & -p^{-1}_{22}
\end{bmatrix} 
$$
(where by $p_{ij}^{-1}$ I mean the $i,j$th entry of the inverse matrix) which jointly seem to suggest that $\mathbf{P} = \mathbf{I}$. On the otherhand keeping $\mathbf{P}$ fixed then $c_2$ is determined as soon as $c_1$ is which suggests the tangent space is only one dimensional.
Would be very interested in comments or a solution for the general case.
