# Orthogonal mapping $f$ which preserves angle between $x$ and $f(x)$

Let $f: \mathbf{R}^n \rightarrow \mathbf{R}^n$ be a linear orthogonal mapping such that $\displaystyle\frac{\langle f(x), x\rangle}{\|fx\| \|x\|}=\cos \phi$, where $\phi \in [0, 2 \pi)$. Are there such mapping besides $id$, $-id$ in case whe $n$ is odd? Is it true that if $n=2k$ then then there exists an orthogonal basis $e_1,\ldots,e_{2k}$ in $\mathbf{R}^{2k}$ such that the matrix $F$ of $f$ in that basis is of the form $$F=\left [ \begin{array}{rrrrrrrr} A_1 & & &\\ & A_2 & &\\ & & \ddots \\ & & & A_{k}\\ \end{array} \right ],$$ where $$A_1=A_2=\cdots=A_{k}=\left [ \begin{array}{rr} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \\ \end{array} \right ] ?$$

Thanks.

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$F$ as you've given it is a $4k \times 4k$ matrix. I think just leaving out the $B_i$ blocks should fix it, though. –  Ilmari Karonen Aug 30 '11 at 18:03
Yes, you have right. –  Richard Aug 30 '11 at 20:22

Clearly, if $\cos\phi=\pm1$, we must have $F=\pm id$. Thus we may consider only the case $\cos\phi\not=\pm1$.

A real normal matrix $A$ can be orthogonally block-diagonalized. More specifically, there exists orthogonal matrix $Q$ such that $QAQ^T = (|\lambda_1|R(\theta_1)) \oplus\ldots\oplus(|\lambda_k|R(\theta_k))\oplus \Lambda$, where each $\lambda_i$ is some (perhaps complex) eigenvalue of $A$, $R(\theta)$ means the $2\times 2$ rotation matrix by angle $\theta$, and $\Lambda$ is a real diagonal matrix. See, for instance, Theorem 2.5.8 (p.105) of Horn and Johnson's Matrix Analysis.

Now, if $A$ is orthogonal, we must have $|\lambda_1|=\ldots=|\lambda_k|=1$ and all diagonal entries of $\Lambda$ must be $\pm1$. In other words, for some real orthogonal matrix $Q$, we have $F = QAQ^T = R(\theta_1)\oplus\ldots\oplus R(\theta_k)\oplus I_p\oplus (-I_q)$ for some $k,p,q$. Now, if $\langle Fx, x\rangle = \|Fx\| \|x\|\cos \phi$ for all $x$, similar equalities must also hold blockwise. Therefore, the two blocks $I_p$ and $-I_q$ must be void and $\theta_1,\ldots,\theta_k=\pm\phi$. So your assertion is correct. However, note that $\begin{pmatrix}0&1\\1&0\end{pmatrix}R(-\phi)\begin{pmatrix}0&1\\1&0\end{pmatrix}^T = R(\phi)$. Therefore, we can actually choose an orthonormal basis such the matrix representation of $f$ is $R(\phi)\oplus\ldots\oplus R(\phi)$.

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Thanks. Very nice answer. –  Richard Aug 30 '11 at 20:25

Your orthogonal transformation $f$ can be "complexified" to a unitary transformation $U$ on ${\mathbb C}^n$ such that $U(x + i y) = f(x) + i f(y)$ for $x, y \in {\mathbb R}^n$. Being normal, $U$ can be diagonalized, and its eigenvalues have absolute value 1.

The only possible real eigenvalues are $\pm 1$, in which case the eigenvectors can be chosen to be real as well; this corresponds to $\cos \phi = \pm 1$, and your cases $id$ and $-id$.

If $v = x + i y$ is an eigenvector for a non-real eigenvalue $\lambda = a + b i$, then $\overline{v} = x - i y$ is an eigenvector for $\overline{\lambda} = a - b i$. Thus the eigenspaces for complex eigenvalues pair up.
Now $Uv = \lambda v = (a x - b y) + i (a y + b x)$, so $f(x) = a x - b y$ and $f(y) = a y + b x$. Then $x^T f(x) = a \|x\|^2 - b x^T y = (\cos \phi) \|x\|^2$ and $y^T f(y) = a \|y\|^2 + b x^T y = (\cos \phi) \|y\|^2$. Therefore $b x^T y = (a - \cos \phi) \|x\|^2 = -(a - \cos \phi) \|y\|^2$. Since $b \ne 0$, $\|x\|>0$ and $\|y\|>0$, we must have $x^T y = 0$ and $a = \cos \phi$. Since $|\lambda\|=\sqrt{a^2 + b^2} = 1$, $b = \pm \sin \phi$. This gives you your $2 \times 2$ block $\pmatrix{\cos \phi & -\sin \phi \cr \sin \phi & \cos \phi\cr}$ or $\pmatrix{\cos \phi & \sin \phi \cr -\sin \phi & \cos \phi\cr}$ (but you don't need both of them).

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Thanks for your beautiful answer. –  Richard Aug 30 '11 at 21:43