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Prove that, if $T$ is an orthogonal transformation on $\mathbb R^2$ such that $\det T = -1$, there exists an orthonormal basis for $\mathbb R^2$ such that the matrix of $T$ with respect to this basis is

$$ \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}. $$

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What have you tried? Do you know what an orthogonal transformation is? If yes, can you please include the definition in your question? Thank you. – Matt N. Oct 14 '12 at 20:01
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What properties do orthogonal matrices have with regards to diagonalization? What must the diagonal entries be? Why can't T be a rotation? – wj32 Oct 14 '12 at 20:52

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The transformation is a reflection. There is a vector $e_1$ such that $Te_1=-e_1$ and a vector $e_2$ such that $Te_2 =e_2$. Choose $e_1, e_2$ as your basis.

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But of course OP has to prove all the assertions in this answer. – Gerry Myerson Oct 15 '12 at 0:54
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Yes. He has to do the other half of the work. – Pantelis Damianou Oct 15 '12 at 20:39
If $A$ is an orthogonal transformation and $\lambda$ an eigenvalue of $A$ then $\lambda$ has modulus $1$. In fact $$(v,v)=(Av, Av)=(\lambda v , \lambda v )=|\lambda|^2 (v,v) \ .$$ Therefore $|\lambda |=1$. – Pantelis Damianou Oct 17 '12 at 19:46
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$T$ is an unitary transformation, so it's diagonalizable. And because it's over $\mathbb{R}^2$ and $\det(T)=-1$, its eigenvalues are 1 and -1. So it can be diagonlized into $diag(-1,1)$.

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It is not true in general that orthogonal transformations have real eigenvalues. – Pantelis Damianou Oct 16 '12 at 9:57

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