The question is taken from PSet 6.4, Q27 of Strang's Introduction to Linear Algebra, 5th edition.
Find all 2 by 2 matrices that are orthogonal and also symmetric. Which two numbers can be eigenvalues of those two matrices?
I have assumed in what follows that he also meant to say real matrices, as I can find examples not included in the answer that are complex and also satisfy the conditions.
The answer states that orthogonality and symmetry mean that $|\lambda| = 1$, (Which is only true if the matrix is orthogonal and real, I think?) and that $\lambda$ must be real (Since the matrix is symmetric and real), so $\lambda = \pm 1$.
He then goes on to say that the matrix must be either $\pm I$, or
$
S = Q\Lambda Q^T =
\begin{bmatrix}
\cos{\theta} & -\sin{\theta}\\
\sin{\theta} & \cos{\theta}
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}
\begin{bmatrix}
\cos{\theta} & \sin{\theta}\\
-\sin{\theta} & \cos{\theta}
\end{bmatrix}
= \begin{bmatrix}
\cos{2\theta} & \sin{2\theta}\\
\sin{2\theta} & -\cos{2\theta}
\end{bmatrix}
$.
This seems to come from nowhere to me. It's clear that a matrix with such properties must have the form $S = Q\Lambda Q^T$, but the rotation matrix as the eigenvector matrix doesn't seem to have a motivation to me other than the fact that it just works.
I'm also not at all sure what the justification is for saying this is the only form such a matrix can take. I can't see anything wrong with it, so I'm sure this describes at least some of the required matrices, but why is it the case that it describes them all?
Best Answer
Orthogonal matrices are meant to be real matrices; their complex counterparts are known as unitary matrices.
Now, the only orthogonal $2×2$ matrices possible are $$\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$ and $$\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix}$$ This you can prove by assuming the matrix to be $$\begin{bmatrix} a & b \\ c & d\end{bmatrix}$$ Then applying the orthogonality condition ,i.e. $AA^T=I$
Hope it helps:)