Let $u$ and $v$ be orthogonal 3d unit vectors. Let $w = u \times v$ and
$$ A=\begin{bmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{bmatrix}$$ Which of following are correct?
a. $A^{-1} = A^T$
b. $\det (A) = 1$
c. The spectrum of $A$ contain $1$ or $-1$
Choices are {a} or {a,b} or {a,b,c}.
About (a), $A^T$ is orthogonal, so $AA^T=I$. thus $A^{-1}=A^T$. TRUE
About (b), det(A) is 1 or -1. FALSE
What confuses me is (c).
$AA^T=I$, so $AA^T$ has a eigenvalue $\lambda_i$ of 1. thus $\sqrt{\lambda_i}=1$ which leads to $|\lambda_i|=1$.
Could $|\lambda_i|$ not contain 1 or -1 which means (c) is FALSE?
No, the eigenvalues of an $n\times n$ orthogonal matrix, where $n>2$, must be of absolute value $1$, because if $Ax=\lambda x$, then $\|x\|=\|Ax\|=|\lambda|\|x\|$, hence $|\lambda|= 1$. ($\|x\|=\|Ax\|$ follows from the orthogonality of $A$). When $n=3$ there is always a real eigenvalue, so it must be $\pm 1$. Other eigenvalues can be complex valued, in which case they need not be $\pm 1$, but their absolute value is still $1$, which means that they lie on the boundary of the unit disk in the complex plane, that is, they have the form $e^{i\theta}$ for some real $\theta$.
For $n=2$ it is possible that no eigenvalues exist. The geometry of the real plane allows rotations which keep no direction fixed. For example, the map sending $(x,y)$ to $\frac{1}{\sqrt{2}}(x-y,x+y)$ is an orthogonal transformation that has no eigenvalues, because there is not even one direction that is not rotated away.