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Let $M_{2}(\mathbb{R})$ be the ring of $2\times 2$ matrices over the reals and $M_{2}(\mathbb{R})^*$ the set of invertible such matrices. Consider any $A \in M_{2}(\mathbb{R})$ such that $ A^{2}=-I$, where $I$ is the identity element in $M_{2}(\mathbb{R})$. We say that $A,B \in M_{2}(\mathbb{R})$ are conjugate in $M_{2}(\mathbb{R})$ if there exists $C \in M_{2}(\mathbb{R})^*$ with $B=C^{-1}AC$.

What is the conjugacy class of $A$?

My main problem is that $S:=\{A \in M_{2}(\mathbb{R})$: $ A^{2}=-I\}$ is neither a subring nor an ideal so I cannot use any "structural result" to show what the conjugacy class is.

I tried the classic and sometimes efficient "brutal force" approach and calculated that the matrices in $S$ are of the form $ \left( \begin{array}{ccc} a & b \\ -\frac{a^2+1}{b} & -a \ \end{array} \right)$, with $b \neq 0$. Then, I conjugated any of these matrices by an arbitrary invertible matrix but this did not lead anywhere.

Any idea as to what the conjugacy class should be?


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up vote 2 down vote accepted

You already got it! Try to prove that it is $S$ itself.

Hint: It would be enough to prove that for any $A\in S$, there is a basis, according to which the matrix of $v\mapsto Av$ is $\pmatrix{0&1\\-1&0}$. I think, you can start out of any nonzero vector $v_1\in\Bbb R^2$, maybe the easiest to consider $v_1:=\pmatrix{0\\1}$. Now $v_2$ is given by $[A]_{v_1,v_2}$.

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I had a feeling it was $S$ itself! At least now I know what I have to do. Thanks! –  dado Feb 23 '13 at 12:42
I had doubts, but then checked, and it works... Have you known that, putting the basis vectors in the culomns to a matrix $B$ then $[A]_B=B^{-1}AB$? That's why we look for such a basis. –  Berci Feb 23 '13 at 13:08
What do you mean by $[A]_B$? –  dado Feb 23 '13 at 13:17
The matrix of $v\mapsto Av$ w.r.t. the basis given by $B$. –  Berci Feb 23 '13 at 13:23
Ok. Now it makes sense. Very clever idea, thank you! –  dado Feb 23 '13 at 13:53
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