If $\lambda$ is an eigen value of an orthogonal matrix $A$, then show that $\frac{1}{\lambda}$ is also an eigen value of $A$ If $\lambda$ is an eigen value of an orthogonal matrix $A$, then show that $\frac{1}{\lambda}$ is also an eigen value of $A$, with the same set of eigen vectors.
I have proceeded like this:
If $A$ is orthogonal, $A^{-1}=A^{T}$
So, the characteristics equation : $0=|A-\lambda I_n|=|A^T - \lambda I_n|=|A^{-1} - \lambda I_n|=|A^{-1}(I_n-\lambda A)|=|A^{-1}|(-\lambda)^n|A-\frac{1}{\lambda}I_n| \implies |A-\frac{1}{\lambda} I_n|=0$
So, $\frac{1}{\lambda}$ is also an eigen value.
How can I show that it has the same set of eigen vectors?
 A: $A$ is orthogonal if and only if $A^{-1}=A^{\top}$.
Suppose $\mathbf{v}\ne\mathbf{0}$ is an eigenvector of some orthogonal marix $A$ associated with its eigenvalue $\lambda$. Since $A$ is invertible, $\lambda\ne 0$. We have
$$
A\mathbf{v}=\lambda\mathbf{v}.
$$
Multiply $A^{\top}$ on both sides, and
$$
\mathbf{v}=A^{\top}A\mathbf{v}=\lambda A^{\top}\mathbf{v}.
$$
Since $\lambda\ne 0$, this implies that
$$
A^{\top}\mathbf{v}=\frac{1}{\lambda}\mathbf{v}.
$$
Since $\mathbf{v}\ne\mathbf{0}$, the above equation indicates that $\mathbf{v}$ is also an eigenvector of $A^{\top}$ associated with an eigenvalue $1/\lambda$.
Using this trick, in general, any eigenvector of an orthogonal matrix $A$ is also an eigenvector of $A^{\top}$, and vice versa. Therefore, an orthogonal matrix shares all of their eigenvectors with its transpose/inverse.
A: Suppose $~λ~$ is an eigen value of $~A~$.
If $~A~$ is orthogonal, $~A^{-1}= A'~\tag1$
Now, eigen values of $~A~$ and $~A'~$ are always same.
Also, eigen values of $~A^{-1}=\frac{1}{λ}~$
From $(1)$ eigen values of $~A' =~$ eigen values of $~A^{-1}~$
Thus, $~\frac{1}{λ}~$ is an eigen values of $~A'~$ and also of $~A~$.
