I'm reading through G&P's Differential Topology book, but I hit a wall at the end of section 4. There is a result stating
The set $X=\{A\in M_{m\times n}(\mathbb{R}):\mathrm{rk}(A)=r\}$ is a submanifold of $\mathbb{R}^{m\times n}$ with codimension $(m-r)(n-r)$.
There is a suggestion: Let $A\in M_{m\times n}(\mathbb{R})$ have form
$$
A=\begin{pmatrix} B & C \\ D & E\end{pmatrix}
$$
where $B$ is an invertible $r\times r$ matrix. Then right multiply by
$$
\begin{pmatrix} I & -BC^{-1} \\ 0 & I \end{pmatrix}
$$
and show $\mathrm{rk}(A)=r$ iff $E-DB^{-1}C=0$.
I multiplied out and got the matrix
$$
M:=\begin{pmatrix} B & 0 \\ D & E-DB^{-1}C\end{pmatrix}.
$$
Since I multiplied by a nonsingular matrix, I know that $\mathrm{rk}(A)=\mathrm{rk}(M)$. If $E-DB^{-1}C=0$, then
$$
M=\begin{pmatrix} B & 0 \\ D & 0 \end{pmatrix}
$$
has rank $r$, so $A$ has rank $r$. For the converse, if $A$ has rank $r$, then $M$ has rank $r$, so by performing row operations, $M$ is row equivalent to a matrix of the form
$$
\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}.
$$
This would imply $E-DB^{-1}C$ is row equivalent to $0$, and I think this implies $E-DB^{-1}C=0$.
My main concern is then, how does this approach imply $\mathrm{codim}(X)=(m-r)(n-r)$? Is there some special map I can apply the Preimage Theorem to?
Thank you.
Best Answer
Since you asked, I've replaced my hint with a full solution:
First consider matrices of the form
$$A = \begin{pmatrix} B & C\\ D & E \end{pmatrix}$$
Where $B$ is an $r \times r$ nonsingular matrix. Since invertibility is an open condition, this set of such matrices, denoted $Z$, is a submanifold of $M_{m \times n}$. Postmultiply by the nonsingular matrix
$$\begin{pmatrix} I & -B^{-1}C\\ 0 & I \end{pmatrix}$$
to obtain the matrix
$$\begin{pmatrix} B & 0\\ D & -DB^{-1}C + E \end{pmatrix}$$
the original matrix has rank $r$ iff this new matrix has rank $r$, which is clearly only the case if $-DB^{-1}C + E = 0$. Thus we can define a map $f$ from $Z$ to matrices of size $(m-r) \times (n-r)$ that sends $A$ as above to $-DB^{-1}C + E$. This is clearly smooth, so it suffices to check that it is a submersion. Now, the tangent space of the image is the same space as the image, since the image is a linear space. Let $X$ be an $(m-r) \times (n-r)$ matrix. Consider the curve passing through any matrix $A \in Z$.
$$\gamma(t) = \begin{pmatrix} B & C\\ D & E+tX \end{pmatrix}$$
The derivative of $f \circ \gamma$ at $0$ is $X$, and this is equal to
$$df_{A}(\begin{pmatrix} 0 & 0\\ 0 & X \end{pmatrix})$$
so that at any arbitrary point $A$ we have shown the existence of a tangent vector at $A$ that is mapped by $df$ to $X$. This verifies that $f$ is submersion, and hence $f^{-1}(0)$ is a smooth submanifold of $\mathbb{R}^{mn}$. The dimension $f^{-1}(0)$ is $mn - (m-r)(n-r)$, i.e. of codimension $(m-r)(n-r)$.
Of course, we have only shown that matrices of rank $r$ contained in $Z$ form a smooth submanifold. However, any matrix can be put into the form of matrices in $Z$ by rearranging rows and columns, which is just a linear isomorphism. Thus if $A$ is matrix of rank $r$, we have a map $R$ to a matrix in $Z$ contained in chart $\psi$. Then we have that $\psi \circ R$ is a smooth chart around $A$ inherited from a chart on $M_{m \times n}$. The collection of these charts then extends to a maximal atlas giving the set of rank-$r$ matrices the structure of a smooth submanifold.