# Angle preserving linear maps [duplicate]

In Spivak's Calculus On Manifolds, in part (c) of question 1-8, he asks the following question: What are all angle preserving $T:\mathbf{R}^n \to \mathbf{R}^n$?

I already showed that if $T$ is diagonalizable with a basis $\{x_1,\ldots,x_n\}$ where $Tx_i = \lambda_i$, then $T$ is angle preserving $\iff$ $|\lambda_i| = |\lambda_j|$ for all $i,j$ (this was part (b)). Perhaps this can be used?

Thanks.

## marked as duplicate by user1551, Amzoti, Chris Godsil, Martin Argerami, Mark BennetJun 7 '13 at 3:22

• Is $T$ a linear function? Otherwise conformal maps are also among the solutions. – Mario Carneiro Jun 7 '13 at 0:30
• This isn't specified in the book, but I believe we take $T$ to be linear (just from the context). – nigel Jun 7 '13 at 0:32
• Then the solution should be the orthogonal matrices, i.e. $T\in O(n)$. Have you tried exploring if you can prove $T^\dagger T=I$? – Mario Carneiro Jun 7 '13 at 0:33
• Mario: Won't it be the matricies $\lambda M$ where $\lambda$ is any number and $M$ is an orthogonal matrix? Even then, I'm not sure this is solution the author is looking for. – nigel Jun 7 '13 at 0:37
• In fact, what I just said in the above comment (the solutions being $\lambda M$ where $M$ is orthogonal) only seems to work when $T$ is diagonalizable. – nigel Jun 7 '13 at 0:46

Take any orthonormal basis $\mathbf e_1,\cdots,\mathbf e_n$. These are mapped to a set $T \mathbf e_j = \mathbf f_j$ which are also orthogonal (and assumed non-zero). Hence in our orthonormal basis, $T$ is a matrix with columns $\mathbf f_j$ which are orthogonal. $D=T^\dagger T$ is a diagonal matrix with entry $(jj)$ given by $\lVert \mathbf f_j\rVert^2$. Further, considering the angle between $\mathbf e_1$ and $\mathbf e_1+\mathbf e_k$ gives

$$\cos \theta = 1/\sqrt 2 = \frac{\mathbf e_1^\dagger D (\mathbf e_1+\mathbf e_k)}{\sqrt{\mathbf e_1^\dagger D \mathbf e_1\times (\mathbf e_1+\mathbf e_k)^\dagger D (\mathbf e_1+\mathbf e_k)}} = \frac{\lVert \mathbf f_1\rVert^2}{\lVert \mathbf f_1\rVert\sqrt{\lVert \mathbf f_1\rVert^2+\lVert \mathbf f_k\rVert^2}}$$ and hence $$\lVert \mathbf f_k\rVert = \lVert \mathbf f_1\rVert$$

Therefore, $D=\lambda I$ is a positive multiple of the identity matrix. Accordingly, $\lambda^{-1/2}T$ is an orthogonal matrix.

Thus since clearly such matrices always preserve angles, we are done. The answer is: non-zero multiples of orthogonal matrices.

Note: Whilst $T$ need not be diagonalizable, $T^\dagger T$ always is. (It's a real, symmetric matrix.)

• If I am not mistaken, you are using the fact that $D=T^\dagger T$ is angle preserving. This would be true if we knew $T^\dagger$ was angle preserving, but I do not see how to deduce this without knowing a-priori that $T$ is a multiplication of an orthogonal matrix. – Asaf Shachar Nov 29 '16 at 11:44
• @AsafShachar It's been a while(!) but I don't think I use any such property; the statement that $T$ preserves angles is the statement that $\left< T \mathbf{u}, T \mathbf{v}\right>/\lVert T\mathbf{u} \rVert\lVert T\mathbf{v} \rVert = \left< \mathbf{u}, \mathbf{v}\right>/\lVert \mathbf{u} \rVert\lVert \mathbf{v} \rVert$. Expressing these inner products in terms of $T^\dagger$ gives the second equality in the display formula. – Sharkos Nov 29 '16 at 20:04
• Thanks! I was indeed confused about something. Your proof is very nice. – Asaf Shachar Nov 29 '16 at 22:07
• @AsafShachar No problem! – Sharkos Nov 29 '16 at 22:14