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$C$ 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?

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    $\begingroup$ It's wrong that if $det(A) = \lambda_1 \lambda_2 \cdots$ then $\lambda_1,\lambda_2 ,\cdots$ are eigenvalue of $A$. If it was true, then every number would be eigenvalue of a non invertible matrix. Indeed, $\forall x, det(A) = 0 = 0 \times x \times \cdots $ $\endgroup$ – Tryss Nov 22 '15 at 21:25
  • $\begingroup$ That the matrix is orthogonal is a bit of a red-herring. It'd be true for the inverse of any (invertible) matrix. $\endgroup$ – quid Nov 22 '15 at 21:26
  • $\begingroup$ @quid The question concerns about the eigenvalues of the transpose, not the inverse. $\endgroup$ – user1551 Nov 22 '15 at 21:57
  • $\begingroup$ @user1551 The matrix is orthogonal so the transpose is the inverse. The assertion is true for any invertible matrix when considering the inverse. If the matrix is orthogonal this just happens to be the transpose. My point is that the proof is rather springs to mind when asked for the inverse. $\endgroup$ – quid Nov 22 '15 at 22:04
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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 $$\frac{1}{\lambda} x = C^Tx.$$

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