# 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?

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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 $T(u)=u$.

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The product of all the non-real eigenvalues, since they come in conjugate pairs, must be positive. Since $T$ is orientation preserving, its determinant is $1$. Thus the product of all the real eigenvalues must be positive as well. In particular, the product of all the "remaining" real eigenvalues must be positive. Since there are an odd number of "remaining" eigenvalues, that means at least one of them must be positive, i.e., there is a $\lambda_u$ that is $1$. – froggie Nov 28 '12 at 16:33

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.)

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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.

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