Does the set of $n$ by $n$ matrices of rank $q$ form a manifold?

Question

I'm not sure whether the space of all rank-$q$ square matrices of dimension $n$ is a submanifold. I have totally no clue. Can somebody help?

Answer 1

It is true. I was unable to complete the ideas in my initial answer, so I offer a new answer, which I admittedly learned from books. But I might as well put it here as it is under interest. Take a matrix of rank $q$, and write it in block form, $$\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ where $A$ is $q\times q$ and of rank $q$. Any such matrix of rank $q$ can be put in this form by exchanging rows, but it is not really necessary, it is more to exposit the proof that we take this form. Now essentially we preform row operations, which is equivalent to multiplying by a matrix, $$\begin{pmatrix}I&0\\-CA^{-1}&I\end{pmatrix}\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&B\\0&D-CA^{-1}B\end{pmatrix}$$ Now it suffices to find a manifold neighbourhood of this last matrix.This matrix will have rank $q$ if and only if $D-CA^{-1}B=0$. Now this is a map from $\mathbb{R}^{n^2}$ to $\mathbb{R}^{(n-q)^2}$ we need to see that the derivative of this map has rank $(n-q)^2$, however the fact that there are the $(n-q)^2$ coordinates of the matrix $D$ there insures this fact. So the zero set of these functions is a submanifold. Incidentally of dimension $n^2-(n-k)^2=k(2n-k)$.