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Let $A$ be a real orthogonal matrix. Then $A^{\text T} A = I.$ Let $\lambda \in \Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X \in \Bbb C^n.$ Then we have

$$\begin{align*} X^{\text T} A^{\text T} A X = X^{\text T} X. \\ \implies (AX)^{\text T} AX & = X^{\text T} X. \\ \implies (\lambda X)^{\text T} \lambda X & = X^{\text T} X. \\ \implies {\lambda}^2 X^{\text T} X & = X^{\text T} X. \\ \implies ({\lambda}^2 - 1) X^{\text T} X & = 0. \end{align*}$$

Since $X$ is an eigenvector $X \neq 0.$ Therefore ${\|X\|_2}^2 = X^{\text T} X \neq 0.$ Hence we must have ${\lambda}^2 - 1 = 0$ i.e. ${\lambda}^2 = 1.$ So $\lambda = \pm 1.$

So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $\pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.

What's going wrong in my argument above. Please help me in this regard.

Thank you very much for your valuable time.

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The mistake is your assumption that $X^TX\ne0$. Consider a simple example: $$A=\pmatrix{0&1\\-1&0}.$$ It is orthogonal, and its eigenvalues are $\pm i$. One eigenvector is $$X=\pmatrix{1\\i}.$$ It satisfies $X^TX=0$.

However, replacing $X^T$ in your argument by $X^H$ (complex conjugate of transpose) will give you the correct conclusion that $|\lambda|^2=1$.

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  • $\begingroup$ how can Euclidean norm of non zero vector be zero? $\endgroup$ – math maniac. Mar 31 at 5:08
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    $\begingroup$ @mathmaniac. How can $1^2+i^2$ equal zero? $\endgroup$ – Lord Shark the Unknown Mar 31 at 5:12
  • $\begingroup$ I think the Euclidean norm of $X \in \Bbb C^n$ is $\sqrt {X^{\text T} \overline X}\ \text {or}\ \sqrt {{\overline X}^{\text T} X},$ not $\sqrt {X^{\text T} X}.$ Am I right? $\endgroup$ – math maniac. Mar 31 at 5:13
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    $\begingroup$ Which is same as $\sqrt {X^{\text H}X},$ as you have rightly pointed out. $\endgroup$ – math maniac. Mar 31 at 5:19

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