# Eigenvalues of a real orthogonal matrix.

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.

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 $$|\lambda|^2=1$$.
• @mathmaniac. How can $1^2+i^2$ equal zero? – Angina Seng Mar 31 '19 at 5:12
• 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? – math maniac. Mar 31 '19 at 5:13
• Which is same as $\sqrt {X^{\text H}X},$ as you have rightly pointed out. – math maniac. Mar 31 '19 at 5:19