# Showing that this is a group under matrix multiplication

I need to prove that $P=\left\{A\in M_2(\mathbb{R})\mid A^TXA = X\right\}$ is a group under matrix multiplication.

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For closure: $(AB)^T X AB = B^T A^T X AB = B^T X B = X$. – Rudy the Reindeer Dec 29 '11 at 11:57
Yes that's enough. – Rudy the Reindeer Dec 29 '11 at 12:04
John, with this kind of an exercise you don't need to worry about associativity and identity properties, because the group will inherit these from the matrix group (where we already know that associativity and identity properties hold), and your group will be a subset of that, so those axioms will automatically hold. You only need to worry about closure, and that's exactly what the subgroup criteria will give you. Concentrate on those, that is the real work! – Jyrki Lahtonen Dec 29 '11 at 12:19
I guess it is worth emphasizing that in this case you didn't limit $A$ to a group of matrices, so you do need to check the existence of an inverse as in matt's answer. Sorry, I gave my knee-jerk response without checking all the details first. – Jyrki Lahtonen Dec 30 '11 at 14:45

Closure: Let $A,B\in P$ then:

$$(AB)^T X(AB) = B^TA^TXAB = B^T(A^TXA)B = B^TXB=X$$

Therefore we have $(AB)\in P$.

Inverse: First observe that $X$ is invertible since $\det(X)=3\times 1-1\times 1 =2\not=0$. Now suppose $A\in P$ then we have:

$$\det(A^TXA)=\det(X) \implies \det(A^T)\det(X)\det(A)=\det(X)\implies \det(A^T)\det(A)=1$$

Since $\det(A^T)=\det(A)$ we have that $\det(A)\not=0$, that is, $A$ is invertible ($A^{-1}$ exists).

Moreover, $A^{-1}\in P$ since: $$A^TXA=X\implies X=(A^T)^{-1}XA^{-1}\implies X=(A^{-1})^TXA^{-1}$$

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Good point showing that A^-1 does exist in the first place for a given A. That's not automatically true. – joshin4colours Dec 29 '11 at 17:02

For closure: $(AB)^T X AB = B^T A^T X AB = B^T X B = X$.

For inverses note that $\det(X) \neq 0$ and hence $\det(A^T X A) = \det(A^T) \det (X) \det(A) \neq 0$ and hence $A$ is invertible and so $(A^{-1})^T = (A^T)^{-1}$.

Use this to show that $A^{-1} \in P$.

Hope this helps.

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@John Yes, it's what you have to show. The second line above shows exactly that. : ) – Rudy the Reindeer Dec 30 '11 at 12:09

Hint: For closure, I'd suggest not to work out things explicitly, but use the fact that $(AB)^T=B^TA^T$. For the inverse, you can use the same property.

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That just shows that the identity is contained in your set. So yes, that's OK. – Raskolnikov Dec 29 '11 at 12:03