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

Question

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 $x^T I x= x \cdot x$?

Answer 1

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 $$x^T I x= x^T x = x \cdot x$$

Answer 2

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 $(A\mathbf{x})^T=\mathbf{x}^TA^T$.

Answer 3

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 $$(Ax)^T(Ax)=x^T A^T Ax=x^T x$$