$C$ is an orthogonal matrix. If $\lambda$ is an eigenvalue of $C$, prove $\frac{1}{\lambda}$ is an eigenvalue of $C^T$.
I know $\lambda$ isn't zero because an orthogonal matrix has determinant $1$ or $-1$, so $\det(C) = 1$ or $-1$. And $\det(C)$ is the product of eigenvalues, so must also be $1$ or $-1$.
I'm pretty sure my proof isn't sufficient. Here's what I've got so far:
$CC^T = I$, therefore $\det(C)\det(C^T)= \det(I) = 1$.
$\det(C^T) = \frac{1}{\det(C)} = \frac{1}{\lambda_{1}\lambda_{2}…} = \frac{1}{\lambda_{1}} \frac{1}{\lambda_{2}}…$
So if lambda is an eigenvalue of $C$, then $\frac{1}{\lambda}$ is an eigenvalue of $C^T$. This is pretty sketchy… Is there a better way to prove this?
Best Answer
Let $x$ be an eigenvector of the orthogonal matrix $C$ corresponding to the eigenvalue $\lambda\neq 0$. Then $$Cx = \lambda x.$$
Applying $C^T$ to both sides from the left, and recalling that $C^TC=I$, we get $$x = \lambda C^T x,$$ or $$\frac{1}{\lambda} x = C^Tx.$$