Linear Algebra (Orthogonal Transformations) Suppose $T:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is a orthogonal transformation that preserves orientation and $n$ is even. Suppose also that there is some vector $v\in\mathbb{R}^n$ such that $T(v)=v$. Let $E$ denote the orthogonal complemente of $v$. Show that there exist some $u\in E$ such that $T(u)=u$.
Is this true?

Note that $E$ is a Euclidean space of odd dimension fixed globally by $T$, and the restriction of $T$ to $E$ is orientation-preserving (since extending by the fixed vector $v$ it is orientation-preserving). So the question really is: does an orientation-preserving orthogonal transformation of an odd-dimensional Euclidean space always have a fixed vector?
The answer is yes. The characteristic polynomial $P$ of (the new, restricted) $T$ is unitary, real and of odd degree $2m+1$, and its (complex) roots lie on the unit circle because $T$ is orthogonal. Factoring $P$ into irreducible factors in $\mathbf R[X]$, the irreducible factors are either of the form $X^2-2\cos\alpha X+1$, $X-1$ or $X+1$. Since $\det T=1$ the constant term of $P$ is $(-1)^{2m+1}=-1$, so there is at least one factor $X-1$, and therefore an eigenvector $u$ for theeigenvalue $1$. (Instead of looking at the constant  term, one can also reason: the product of all roots of $P$ must be $1$, so the number of factors $X+1$ is even, and that of the factors $X-1$ therefore odd, whence nonzero.)
