# Conjecture: For $n\in N, n\neq 2$, there exists an eigenvector for any orthogonal operator T on $R^n$.

(This question is my assumption, I'm not sure if it's true.)

I have already proven cases where $$n$$ is odd. Since the characteristic polynomial $$p_T(t)$$ of $$T$$ on $$R^n$$ is in the form $$p_T(t) = t^n + \dots + \det([T]_\beta)$$ Since $$n$$ is odd, we have the sign of $$\lim_{t\to -\infty}p_T(t)$$ is different from $$\lim_{t\to \infty}p_T(t)$$, which implies that $$p_T(t)$$ has a solution on $$R$$. Thus, $$T$$ has an eigenvalue, which means there exists an eigenvector.

However, I have no idea about cases when $$n$$ is even.

When $$n=2$$, the rotation matrix is a counterexample.

When $$n>2$$, intuitively the orthogonal operator on $$R^n$$ is like rotation on a 2-dimension plane, which kinda implies that exists vector $$v$$ in the remaining $$n-2$$ dimensions, which is not affected by that rotation (Then $$Tv = \pm v$$).

But I can't find any relationship between orthogonal and even dimensions. Can anyone give me a hint?

FYI: I comes up with this question when I'm proving

"Suppose $$S\in \mathcal{L}(R^3)$$ is orthogonal; Prove that $$\exists x\in R^3, x\neq 0, s.t., S^2 x = x$$",

which is an exercise in linear algebra done right.

• Put the dimensions in pairs and do a rotation in each and every pair: Like a matrix with blocks $\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&\phantom{-}1\end{pmatrix}$ in the diagonal.
– user551819
Commented Apr 16, 2018 at 20:09
• @totoro Can you kindly be more specific? Commented Apr 16, 2018 at 20:21
• I just gave you a matrix. Compute its eigenvalues.
– user551819
Commented Apr 16, 2018 at 20:21
• @totoro Are you trying to show that it has no eigenvector? But it's a two dimensional one, which is not considered in this question. Commented Apr 16, 2018 at 20:23
• No, put that as blocks in the main diagonal to form a matrix of an arbitrary even size.
– user551819
Commented Apr 16, 2018 at 20:24

Consider $\begin{pmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0&0\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0&0\\ 0&0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\0&0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}$.