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I am thinking I use that $\lambda$ is an eigenvalue of $A$ iff $\operatorname{det}(\lambda I - A)=0$.

Also I know that for an $n\times{n}$ matrix, there are at most $n$ distinct eigenvalues, so for our $2 \times 2$ matrix, there must be at most $2$ distinct eigenvalues.

Since orthogonal means that $A A ^T = I$...?? ... not sure how to use any of this.
I think part of my problem here is it has been way too long since I took the computational part of linear algebra, so there's probably some core elements I am forgetting.

Any suggestions on how to start tackling this would be appreciated.

2 Answers 2

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Hint: Note that the determinant of any matrix is the product of its eigenvalues. In this case, we necessarily have $\lambda_1\lambda_2 = 1$.

All that's left is to show that the eigenvalues satisfy $|\lambda| = 1$. To this end, note that because $A$ is orthogonal, it satisfies $\|Av\| = \|v\|$ for any vector $v$, including (complex) eigenvectors. Alternatively, note that $$ Av = \lambda v \implies (Av)^*(Av) = (\lambda v)^*(\lambda v) $$ where $*$ here denotes the conjugate transpose.

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    Thank you, I am able to use the transpose to show λ is +/- 1, but I have no reference on how to get to e^i(theta). And from my understanding λ is +/-1 if det (A) = -1, not if det (A) = 1 (I could be wrong though). I know Euler's formula, does it somehow involve that? – 2017-02-03
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    https://www.youtube.com/watch?v=kst2Io91JbM This video sort of helped clarify a few things. I am still a bit confused - shouldn't det A = -1 lead me to the complex answers, and det A =1 have eigenvalues of +/- 1? I will keep working on it, maybe it will all click. – 2017-02-03
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Under the assumption that $A$ is a $2 \times 2$ orthogonal matrix satisfying $\det(A)=1$ one has that $\mbox{adj}(A)=A^T$ is the inverse of $A$. Thus, the entries of $A$ satisfy the set of conditions $a_{11}=a_{22}$ and $a_{12}=-a_{21}$.

Hence, the determinant condition $\det(A)=1$ lead us to the equation $a_{11}^2+a_{12}^2=1$, and whence, A is a matrix generated by the column vectors $v_1=(\cos(\theta),\sin(\theta))^T$ and $v_2=(-\sin(\theta),\cos(\theta))^T$ satisfying the orthogonality condition $v_1^Tv_2=0$. Wih respect to the eigenvalues it is now clear from the above construction that one has the characteristic equation $$ \det(A-\lambda I_2)=\lambda^2-2\lambda \cos(\theta)+1=(\lambda-e^{i\theta})(\lambda-e^{-i\theta}).$$ In case where $\theta\neq k\pi$, the eigenvalues $\lambda_{\pm}=e^{\pm i \theta}$ are complex numbers.