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(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.

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  • $\begingroup$ 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. $\endgroup$
    – user551819
    Commented Apr 16, 2018 at 20:09
  • $\begingroup$ @totoro Can you kindly be more specific? $\endgroup$
    – Wenzel
    Commented Apr 16, 2018 at 20:21
  • $\begingroup$ I just gave you a matrix. Compute its eigenvalues. $\endgroup$
    – user551819
    Commented Apr 16, 2018 at 20:21
  • $\begingroup$ @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. $\endgroup$
    – Wenzel
    Commented Apr 16, 2018 at 20:23
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    $\begingroup$ No, put that as blocks in the main diagonal to form a matrix of an arbitrary even size. $\endgroup$
    – user551819
    Commented Apr 16, 2018 at 20:24

1 Answer 1

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With the help of @totoro, the conjecture has been disproved.

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

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