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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?

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    $\begingroup$ If you have $Ax = \lambda x$, then applying $A^T$ on both sides and recalling that it's the inverse of $A$, we get $x = A^T(\lambda x ) = \lambda A^T x$, or $\frac{1}{\lambda} x = A^T x$. $\endgroup$ – Hayk May 13 '18 at 6:02
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$A$ 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.

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    $\begingroup$ But the fact that $A$ and $A^T$ have the same set of eigen vectors is is not true in general..For example $M= \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right]$ is an eigen vector of $A= \left[ {\begin{array}{cc} 0 & -1 \\ 2 & 3 \\ \end{array} } \right]$ but not of $A^T$ $\endgroup$ – Legend Killer May 13 '18 at 6:16
  • $\begingroup$ Fix the last paragraph, please. It is true that any eigenvalue of $A^T$ is also an eigenvalue of $A$, but that does not hold for eigenvectors. $\endgroup$ – Jyrki Lahtonen May 13 '18 at 6:21
  • $\begingroup$ @LegendKiller: Thank you for your comment! This example is true, yet your $A$ is no longer orthogonal. What I meant is that, as long as $A$ is orthogonal, $A$ and $A^{\top}$ share all of their eigenvectors. I just fixed my statement, and there might be less ambiguity now. $\endgroup$ – hypernova May 13 '18 at 6:55
  • $\begingroup$ @JyrkiLahtonen: Thank you for your comment! There might be some ambiguity. What I meant there was merely for orthogonal matrices. I did not mean for general $A$ and its transpose. I just fixed my statement, and there might be less ambiguity now. $\endgroup$ – hypernova May 13 '18 at 6:56
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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 $~A~$.

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    $\begingroup$ Welcome to MSE. Your answer adds nothing new to the already existing answer. $\endgroup$ – José Carlos Santos Jul 26 at 10:21

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