- Let $S$ be the set $\mathbb R^{m\times n}$ of real $m\times n$ matrices, and let $G=\text{GL}_m(\mathbb R)\times\text{GL}_n(\mathbb R)$. Prove that the rule $(P,Q)\ast A= PAQ^{-1}$ defines an operation of $G$ on $S$.
- Describe the decomposition of $S$ into $G$-orbits.
- Assume $m\le n$. What is the stabilizer of the matrix $[I |0]$?
1 is a straightforward computation. For 2, since any matrix is similar to a matrix the upper left corner of which is the identity sub-matrix, the orbits are characterized by the number of ones in that sub-matrix, i.e., by rank.
The only question is how to describe the stabilizer of $A=[I |0]$? I tried to write the equation $PA=AQ$ and expand the products, but I don't know how to describe the pairs of matrices $(P,Q)$ for which this is true.
If you want to have $P[I|0] = [I|0]Q$, you need to have $$[P|0] = [I|0]Q$$ To work with the right side, split $Q$ into blocks to make sense of the product, that is, put $$Q = \begin{bmatrix} Q_{11} & Q_{12} \\ Q_{21} & Q_{22} \end{bmatrix}$$ where $Q_{11}$ is $m \times m$ and $Q_{22}$ is $(n-m) \times (n-m)$.
This gives us $$\begin{bmatrix} I&0\end{bmatrix}Q = \begin{bmatrix} I&0\end{bmatrix}\begin{bmatrix}Q_{11}&Q_{12}\\Q_{21}&Q_{22}\end{bmatrix} = \begin{bmatrix}Q_{11} & Q_{12}\end{bmatrix}.$$ From this we see that $Q_{11} = P$, $Q_{12} = 0$, and, since $Q$ needs to be invertible, $Q_{22} \in \mathrm{GL}_{n-m}(\mathbb{R})$. This allows us to choose $Q_{21}$ in any way we like.