TITLE: If $\lambda$ is the eigen-value of a $n\times n$ non-singular orthogonal matrix $A$, then prove that $\frac{1}{\lambda}$ is also an eigen-value.
QUESTION [5 upvotes]: QUESTION:
If $\lambda$ is the eigen-value of a $n\times n$ non-singular matrix $A$ and $A$ is a real
  orthogonal matrix, then prove that $\frac{1}{\lambda}$ is an
  eigen-value of the matrix $A$.

MY ATTEMPT:
Since $\lambda$ is the eigen-value of a $n\times n$ matrix $A$, we have 
$$|A-\lambda I_n|=0$$
Also since $A$ is a real orthogonal matrix,we have 
$$AA^T=A^TA=I_n$$
So we can conclude that $$|A-\lambda( AA^T)|=0$$ Or, $$|\lambda A\left(\frac{1}{\lambda}I_n- A^T\right)|=0$$ Or, $$\left|\lambda A\right|\cdot \left|\frac{1}{\lambda}I_n- A^T\right|=0$$ Or, since $A$ is non-singular, $$\left|A^T-\frac{1}{\lambda} I_n\right|=0$$
So, we can conclude that $\frac{1}{\lambda}$ is an eigen-value of the matrix $A^T$.
But how do I prove that $\frac{1}{\lambda}$ is an eigen-value of the matrix $A$?
Is my working faulty? Or is there a mistake in the question?
Please help.

REPLY [2 votes]: Let's prove that for any matrix $A$ we have $\det{A}=\det{A^T}$. Once this is done it is obvious that
$$\det{(A-\lambda\cdot I)^T}=\det{(A^T-\lambda\cdot I)}$$
and this proves that $A$ and $A^T$ have the same characteristic polynomial and therefore the same eigenvalues.
Let's now start wit the Leibniz formula for a determinant
$$\det{A}=\sum_{\sigma\in\Sigma_n}\epsilon(\sigma)a_{\sigma(1),1}\cdots  a_{\sigma(n),n}$$
Transposing we get
$$\det{A^T}=\sum_{\sigma\in\Sigma_n}\epsilon(\sigma)a_{1,\sigma(1)}\cdots  a_{n,\sigma(n)}$$
Because all the $\sigma$ are bijections
$$\det{A^T}=\sum_{\sigma\in\Sigma_n}\epsilon(\sigma)a_{\sigma^{-1}(1),1}\cdots  a_{\sigma^{-1}(n),n}$$
Now reindexing with $\tau=\sigma^{-1}$ we get
$$\det{A^T}=\sum_{\tau\in\Sigma_n}\epsilon(\tau)a_{\tau(1),1}\cdots  a_{\tau(n),n}=\det{A}$$