# Tangent Space of matrices with same rank

Let $\mathbb{R}^{(n)}$ denote the space of symmetric matrices $n\times n$. Define $R_{k}=\{X\in \mathbb{R}^{(n)}\mid \operatorname{rank} X = k \}$.

Show that $R_{k}$ is manifold and that the tangent space of $R_{k}$ at $M$ is the space of all matrices of the form $WM+MW^{T}$.

We separate the cases that each submatrix $r\times r$ (I not sure about this argument). Example:

Let $\mathcal{A}$ the set of matrices $A\in \mathbb{R}^{(n)}$ such that

$$A= \begin{pmatrix} B & C\\ C^{T} & D \end{pmatrix}$$

Where $B$ is $r\times r$ and nonsingular matrix, it is easy to see that $\mathcal{A}$ is open. We have

$$\begin{pmatrix} B & C\\ C^{T} & D \end{pmatrix} \begin{pmatrix} I & -B^{-1}C\\ 0 & I \end{pmatrix} = \begin{pmatrix} B & 0\\ C^{T} & -C^{T}B^{-1}C + D \end{pmatrix}$$

Then, $A$ has rank $r$ if and only if the right side of the equation above has rank $r$ if and only if $-C^{T}B^{-1}C + D=0$. Now if we define $f\colon \mathcal{A}\to \mathbb{R^{(n-r)}}$, $f(A)=C^{T}B^{-1}C + D$.

We have that $f^{-1}(0)=\mathcal{A}\cap R_{k}$ is smooth manifold. (I can show that $0$ is a regular value). I think that I can complete the argument that $R_{k}$ is smooth manifold. I also know that the tangent space of $R_{k}$ at $M \in \mathcal{A}\cap R_{k}$ is equal to $\operatorname{Ker} Df(M)$, where $Df(M)$ is the derivate of $f$ at $M$, but I have no idea how to show that this is $\{WM+MW^{T}\mid W \in \mathbb{R}^{n^2}\}$.

## 1 Answer

Choosing quadratic forms of signature $(a,k-a, n-k)$ for $a=0,..,k$ we see that the set of quadratic forms (= symmetric matrices) of rank $k$ is a union of k+1 disjoint sets. Let us checks that all this sets are manifolds, and compute the tangents spaces.

Let $M$ be a symmetric matrix ; we think of $M$ as a quadratic form.

Let $F: Gl(n, R)\to S$ given by $F(X)= XMX^t$. The theory of quadratic forms (Gauss reduction) implies that the image of $F_M$ is the set of symmetric matrices with the same rank and signature of $M.$

In other words, every connected component of your set is the orbit of $M$ under the action of the Lie group $GL(n,R)$ acting on the set of quadratic forms. It is therefore an immersed manifold. (constant rank theorem).

In order to compute the tangent space, we compute the derivative at the identity $F(Id+H)= M+HM+MH^t+o(H)$, and the image of $F'(Id)$ is exactly your space.

In order to prove that your manifold is in fact a submanifold (not an immersed submanifold), we check that it is locally closed, i.e. there exists a neighborhood $O$ of $M$ such that $R_k\cap O$ is closed in $O.$ This follows from the fact that a matrix is of rank $k$ iff all its $k+1$ minors are zero (closed condition) and some $k$ minor is not zero (open condition). In a neighborhood $O$ of $M$ some $r$ minor is not zero (the same as $M$), and the result follows.