# How to find the tangent space to a matrix space

I have a hard time approaching these types of problems. In an article it had claimed that the tangent space to all symmetric matrices with the same signature as $M$ at a matrix $M$ is the set of all the matrices of the form $WM+MW^T$. So, I started looking for similar and maybe simpler problems of this type, like finding the tangent space to the set of all invertible matrices at $I$, or the tangent space to the set of all matrices with determinant equal to $1$ at $I$.

Without giving a solution could you give me a hint on how do you start solving such problems? What I know is that I have to find a path on these manifolds which pass through that certain matrix and find the derivative of that path. I've also heard of using the exponential map, but I'm not sure that I've understood it. What is not clear to me is that how do you think of a path for such a manifold, and more importantly, how can one show that this is a manifold?

To make the question clear here is what I want to show: Let $SL_n\mathbb(R)$ be the set of all real $n\times n$ matrices with determinant equal to 1. Show this set is a manifold, and find the tangent space to this manifold at identity.

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Loosely, since the determinant is one scalar constraint, I would expect to find a space of dimension $n^2-1$. Maybe try looking at conditions on matrices $\Delta$ such that $\det(I+\Delta) = 1$? To first order, this is the set of matrices such that $\mathbb{tr} \Delta = 0$. – copper.hat May 21 '12 at 16:47

"Hint 1" .You know that $\mathbb{R}^N$ is a manifold, so the set of matrices is a manifold (think coordinatewise). The determinant is a nice function from that into $\mathbb{R}$, and $SL_n\mathbb(R)$ is the preimage of a regular value, hence a manifold.

Hint 2. The tangent space (at p) to a manifold defined by the vanishing of an equation is the nullspace of the derivative of the defining map at $p$; so consider a curve going into $SL_n\mathbb(R)$ (i.e. a function from an small interval into $\mathbb{R}^{n^2}$, such that its image is an $n^2$-uple corresponding to a special linear matrix), compose with the determinant, and compute the derivative of this at the identity.

Hint 3. This should be a map from the space of matrices to $\mathbb{R}$. Work this out to find the trace. So you recover what copper.hat says above.

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Using the hint 1 and what it says in the Giuseppe Negro's link I can show that $SL_n(R)$ is a manifold. (I still have to go back and review the implicit function theorem and probably the inverse function theorem to figure out how does it exactly work). For the hint 2, here the defining vanishing map is $\det(A)-1=0$, and by copper.hat notations the map will be of the form $c(t)=I+\Delta(t)$, for small $t$, and I think since the determinant is continuous the map will remain inside $SL_n(R)$. Now I should look at $\det(I+\Delta(t))$ and take its derivative. How do I take this derivative? – Keivan May 21 '12 at 17:43
OK, I think I got it, it's entry-wise differentiation. I don't know why I was resisting to accept it! – Keivan May 23 '12 at 23:03

I think I've understood how to solve the whole thing for $SL_n(\mathbb{R})$, so I'll add a complete solution now.

$SL_n(\mathbb{R})=\{A_{n\times n}:\det(A)=1\}$. $\mathbb{R}^n$ is a manifold. $M_n(\mathbb{R})$ (entry-wise) is the same as $\mathbb{R}^{n^2}$, hence it is a manifold. There is a smooth function (determinant) from $M_n(\mathbb{R})$ to $\mathbb{R}$, and $SL_n(\mathbb{R})$ is the pre-image of a regular value (value 1). That is, by implicit function theorem $SL_n(\mathbb{R})$ is a manifold.

Now, let $c(t)=e^{tA}, t\in \mathbb{R}$ be a path in this manifold. In order to have $e^{tA}$ inside the manifold, its determinant should be 1, that is $$1=\det(e^{tA})=e^{tr(tA)}.$$ So, $tr(tA)=0$, that is $tr(A)=0$. On the other hand, $c(0)=I$. Now, taking the derivative we have: $c'(t)|_{t=0}=Ae^{tA}|_{t=0}=A$. That is all the matrices of trace zero are tangent to this manifold.

The dimension of $SL_n(\mathbb{R})=n$, and the dimension of $M_n(\mathbb{R})=n^2$, and the dimension of all the matrices of trace zero is $n^2-n$. So, these are all of the tangent vectors.

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I'll write a solution to the first problem later. – Keivan May 23 '12 at 23:35
Did you figure out the the solution to the first problem? – SorTheene Jun 16 '15 at 21:16