# Question about a proof that the eigenvalues of an $n \times n$ orthogonal matrix are $\pm 1$

Suppose A: $$n \times n$$ orthogonal matrix, $$\lambda$$ is an eigenvalue of A and $$x$$ is corresponding eigenvector.

We know that $$Ax = \lambda x$$

Then $$(Ax)^T (Ax) = x^T A^T Ax = (Ax) \cdot (Ax) = |Ax|^2 = \lambda^2 |x|^2$$.

In addition we know that $$(Ax)^T (Ax) = x^T A^T A x = x^T I x= x \cdot x = |x|^2$$.

So $$\lambda^2 = 1$$. Hence $$\lambda = 1,-1$$.

Why can we say that $$x^T A^T Ax = (Ax) \cdot (Ax)$$ and $$x^T I x= x \cdot x$$?

• This is not true at all. Nearly all orthogonal matrices have complex eigenvectors. Try e.g. $\pmatrix{\cos(\theta) & \sin(\theta)\cr -\sin(\theta) & \cos(\theta)}$. Feb 25, 2019 at 13:21
• @RobertIsrael : this seems to be about real eigenvalues Feb 25, 2019 at 13:22
• Idd my bad, It was supposed to be only about real eigenvalues and not complex eigenvectors. Feb 25, 2019 at 13:23

The standard scalar product on $$\mathbb{R}^n$$ is defines as follows. Let $$\mathbf{x},\mathbf{y}\in \mathbb{R}^n$$, then the standard scalar product between $$\mathbf{x}$$ and $$\mathbf{y}$$ is $$\mathbf{x}\cdot \mathbf{y} := \mathbf{x}^T\mathbf{y}$$. This should answer to both of your questions, together with the observation that $$(A\mathbf{x})^T=\mathbf{x}^TA^T$$.
First you can say $$x^T A^T Ax = (Ax) \cdot (Ax)$$ because $$x^T A^T Ax = (Ax)^T (Ax)$$ and because $$x^Ty = x \cdot y$$ (inner product is the scalar product) Also note that $$Ix = x$$ for any $$x$$, so again $$x^T I x= x^T x = x \cdot x$$
We know $$Ax$$ is a vector, not a matrix. Therefore the inner product of the vector $$Ax$$ tells us $$(Ax)^T(Ax)=(Ax)\cdot(Ax)$$ Any orthogonal matrix $$Q$$ has the property of $$Q^T Q=I$$, so $$(Ax)^T(Ax)=x^T A^T Ax=x^T x$$