Matrices and Linear Algebra Let A be a real $n × n$ orthogonal matrix. Let $X$ be a complex eigenvector of
A with complex eigenvalue λ.
Prove that $X^TX$ = $0$.
Write $X = R + Si$ where R and S are real
vectors. Prove that W spanned by R and S is A-invariant and describe the restriction of A to
W.
Prove that there exists a real orthogonal matrix P such that $P^tAP$ is a block diagonal matrix
with each block of size 1 × 1 or 2 × 2.
 A: Regarding the $\textbf{first claim}$ note $$ X^T X = X^TA^TAX = (AX)^T(AX) = \lambda^2 X^TX $$ which implies $\lambda \in \{ 1, -1\}$ or $X^TX = 0$. But you assume $\lambda$ to be complex.
Regarding the $\textbf{second claim}$ note $AR + iAS = AX = \lambda (R+iS)$ which implies
$$ AR = \operatorname{Re}(\lambda(R+iS)) =  \operatorname{Re}(\lambda)R + \operatorname{Re}(i\lambda)S \\ AS = \operatorname{Im}(\lambda(R+iS)) =  \operatorname{Im}(\lambda)R + \operatorname{Im}(i\lambda)S$$
where $\operatorname{Re}$ denotes the real part $\operatorname{Im}$ respectively. But this means that $AR$ and $AS$ are in $W$. Hence $W$ is $A$-invariant.
For the $\textbf{last claim}$ we note that the first part implies that $R$ and $S$ are orthogonal because $$0 = X^TX = \lVert R \rVert_2^2 + 2i R^TS - \lVert S \rVert_2^2$$
but this means $R^TS=0$ as there is no imagniary part on the left side. Now we can complete $\{R,S\}$ to be basis of $\mathbb{R}^n$. Denote it by $\{R,S,V_3,...,V_n\}$ and define a new matrix $B$ by
$$ Av_j = \sum^n_{i=1} B_{ij} v_i. $$
$B$ has the form
\begin{pmatrix}
* & * \\
0 & * 
\end{pmatrix}
where the block in the upper left corner has dimension 2x2 and the lower right block is zero because $W$ is $A$-invariant. We repeat this procedure for all eigenvalues where the non-complex eigenvalues lead to the 1x1 blocks.
The existence of the $P$ part basically follows from the fact that we are doing a change of basis between orthogonal bases.
