I have some questions on group actions, which occured from this problem:
We have the group $G=\operatorname{GL}_m(\mathbb{R})\times \operatorname{GL}_n(\mathbb{R})$ and the set $X=\operatorname{Mat}_{m\times n}(\mathbb{R})$
The group action is given as follows
$G\times X\to X$, $((P,Q),A)=PAQ^{-1}$.
As described in the text my questions are: How do you find the orbits of the group action in this specific case, and how to show that this actions is faithful or not.
I struggle a little bit with the notation/definition of group actions, as made clear in the text, and feel the need of some explanation.
Thanks in advance for helping me out.
This is indeed a group action as
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$1X=((\mathbb{1}_m,\mathbb{1}_n), X)=\mathbb{1}_mX\mathbb{1}_n=X$
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For $P=(P_1, P_2)$ and $Q=(Q_1, Q_2)$ in $G$ we have $PQ=(P_1Q_1, P_2Q_2)$ and
$(PQ)A=P(QA)$ is immediate to show.
I am asked to describe the orbits of the group action.
In general the orbit of an element $x\in X$ is the set $Gx=\{gx|g\in G\}$.
Also it is obvious that the orbit of the null matrix $0$ is $G0=\{0\}$.
Since the orbits are pairwise disjoint and the union is $X$ we can already conclude that the action is not transitive.
But I struggle to figure out what the other orbits are.
I guess that every other orbit is equal, so we have $GA=GB$ for $A\neq B$ and $A,B$ are not the null matrix.
But I can not proof it yet, as I do not know when you are able to manipulate a matrix $A$ into a matrix $B$ by muliplication with matrices of the general linear group, but I am convinced that this is possible.
I looked in a beginners book about linear algebra, but I did not find something necessary. Also this looks a little bit overly complicated.
However, my main question is about this figuring out if the group action is faithful.
The lecture notes read as follows:
Is $G$ a group and $\varphi$ a homomorphism $\varphi: G\to \operatorname{Sym}(X)$, we say that $G$ acts on $X$ and write $\varphi: G\times X\to X$.
I do not understand.
I know how to get the homomorphism $\varphi: G\to\operatorname{Sym}(X)$, but the group action (for example the one I am working with) is not a homomorphism. In general $X$ is just given as a set, and not with a structure.
To show that a group action is faithful we have to show that $\ker\varphi=\{1\}$.
But with which describtion do I work now? Besides that the kernel is only defined for a homomorphism.
Can I conclude from the group action given, or do I have to transform it into $\varphi: G\to \operatorname{Sym}(X)$?
Do you have to use both describtions side by side?
Best Answer
Edit: I didn't notice that $m$ and $n$ were different.
Let's deal with the case $n=m$ first, as it gives you the right idea for the general case.
Since $P$ and $Q$ are invertible matrices, and you are running over all of them, you may replace $P$ by $PQ$. Then you are looking at $PQAQ^{-1}$. You might as well replace $QAQ^{-1}$ by $A$ because in the end you are likely to want a description that is independent of basis anyway.
So now, you want to find some invariant that distinguishes matrices apart that is not affected by multiplication by an invertible matrix. Once you have done that, then you want to decide if any two matrices with the invariant are related by multiplying on the left and right by invertible matrices, or equivalently as I have shown, a change of basis and a single multiplication.
First, what is the orbit containing $I$?
So once you look at the case $m=n$, you see that you want to use the one matrix to obtain a normal form of some sort, and then you can take transposes to push the 'adjustment' matrix to the other side and take normal forms again. The obvious one to do in general is Gaussian elimination, as that is more or less the only thing we can do with arbitrary matrices.