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:
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.
 A: Suppose that $v\neq 0$ and $Av=\lambda v$. Then,
$$
\frac{1}{\lambda}v=\frac{1}{\lambda}Iv=\frac{1}{\lambda}(A'A)v=\frac{1}{\lambda}A'(Av)=\frac{1}{\lambda}A'\lambda v=A'v.
$$
So $\lambda^{-1}$ is an eigenvalue of $A'$. But $A'$ and $A$ have the same eigenvalues (because $\det(I-\zeta A)=\det[(I-\zeta A)']=\det(I-\zeta A')$ for all real $\zeta$), and so $\lambda^{-1}$ is an eigenvalue of $A$.

Argument that doesn't use determinant: suppose that $v\neq 0$ and $Av=\lambda v$. We have
$$
\lambda^2||v||^2=(\lambda v)'(\lambda v)=(Av)'(Av)=v'A'Av=v'(I)v=||v||^2.
$$
This implies $\lambda^2=1$ and so $\lambda=\pm 1$. In either case, $\lambda=\lambda^{-1}$ and so of course if $\lambda$ is an eigenvalue of $A$ then it's trivial that $\lambda^{-1}$ ($=\lambda$) is also an eigenvalue of $A$.
A: Hints:


*

*if $v$ is eigenvalue for $\lambda$ of an invertible matrix $A$, then it is also eigenvalue for $\lambda^{-1}$ of $A^{-1}$,

*$A$ and $A^T$ have the same characteristic polynomials, hence the same eigenvalues.

*If $A$ is orthogonal then it is invertible, and $A^{-1}=A^T$.
A: 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
$$\det{A^T}=\sum_{\tau\in\Sigma_n}\epsilon(\tau)a_{\tau(1),1}\cdots  a_{\tau(n),n}=\det{A}$$
