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?
 A: 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.)
A: Since $T$ is orthogonal it has a full set of eigenvectors. It is clear also that it has real only elements (otherwise it would be $T:\mathbb{R}^n \rightarrow \mathbb{C}^n$). Therefore all eigenvalues are either real or come in conjugate pairs. $v$ is an eigenvector with eigenvalue that is real, $\lambda =1$. Since the set of remaining eigenvalues for $T$ has size $n-1$ which is odd, there is (at least) one real eigenvalue remaining. Call it $\lambda_u$. From orthogonality it is either $\lambda_u=1$ or $\lambda_u=-1$. I am not exactly sure of what "preserves orientation" means, but I believe that it must mean that $\lambda_u \ne -1$, thus $\lambda_u=1$, and it has vector $u$ such that $T(u)=u$.
A: This should be true, I think. Since $T$ maps $E\cong \mathbb{R}^{n-1}$ into itself, what must be shown is the following:

If $T\colon \mathbb{R}^{n-1}\to \mathbb{R}^{n-1}$ is an orientation preserving orthogonal linear transformation ($n$ even), then there is a $v\in \mathbb{R}^{n-1}$ such that $Tv = v$.

Let's prove this by contradiction. If it were false, then the linear map $T - \mathrm{Id}$ would be invertible. I claim that, in fact, for every $\lambda\in [0,1]$, the linear map $\lambda T - (1-\lambda)\mathrm{Id}$ is invertible. The only way this could be false is if there is a $v\in \mathbb{R}^{n-1}$ such that $\lambda Tv = (1-\lambda)v$, or in other words, that $Tv = \frac{1-\lambda}{\lambda} v$. Since eigenvalues of orthogonal transformations have absolute value $1$, this can only possibly happen when $\lambda = 1/2$. But $(1/2)T - (1/2)\mathrm{Id} = (1/2)(T - \mathrm{Id})$ is invertible since $T - \mathrm{Id}$ is invertible. This proves $\lambda T - (1-\lambda)\mathrm{Id}$ is invertible for all $\lambda\in [0,1]$. Let $\varphi(\lambda) = \det(\lambda T - (1-\lambda)\mathrm{Id})$.  Then $\varphi$ is a continuous nonvanishing function on $[0,1]$. But $\varphi(1) = \det T = 1$ and $\varphi(0) = \det(-\mathrm{Id}) = (-1)^{n-1} = -1$, so $\varphi$ changes sign on $[0,1]$, a contradiction of the fact that $\varphi$ is nonvanishing.
