0
$\begingroup$

From S.L Linear Algebra:

Let $F$ be a rotation through an angle . Show that for any vector $X$ in $\mathbb{R}^3$, we have $||X||=||F(X)||$ (i.e. $F$ preserves norms), where $(a, b)=\sqrt{a^2+b^2}$.


What I tried:

I believe the solution could be found by using a matrix definition of transformation (due to rotation being linear transformation), but I wasn't able to reach the end:

We know that rotation $F(X)=AX$ where $A$ is a standard basis matrix $(a_{ij})$ for column space of $F$.

I believe that dimensions of $A$ should be $3x3$, considering the fact that $A$ is a rotation matrix which is basically an orthogonal matrix with positive determinant (but is $A$ really a rotation matrix?). Considering this, symbolically $AX$ could be represented as:

$$\begin{pmatrix} a_{11} & a_{12} & a_{13 }\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \\ x_3 \\ \end{pmatrix}$$

Hence:

$$||F(X)||=||AX||=\begin{pmatrix} ||a_{11}x_1|| + ||a_{12}x_2|| + ||a_{13}x_3|| \\ ||a_{21}x_1||+||a_{22}x_2|| + ||a_{23}x_3|| \\ ||a_{31}x_1|| + ||a_{32}x_2|| + ||a_{33}x_3|| \\ \end{pmatrix}$$

Therefore, if $A$ is matrix of ones, then $||AX||=||X||$. If steps that I took before were completely valid, how can I show that $A$ is a matrix of ones?

If not, what is $A$ exactly? I know that there is a $2x2$ rotation matrix that contains trigonometric functions as columns (before rotation it obviously contains standard basis), but I couldn't find anything similar for $3x3$ matrix.

Thank you!

$\endgroup$
  • $\begingroup$ You talk about vectors in $\Bbb R^3$, yet write a generic one as $(a,b)$. $\endgroup$ – Lord Shark the Unknown Feb 4 at 19:10
  • $\begingroup$ @LordSharktheUnknown It's not me, it's an author of the book (Serge Lang) that wrote this. Such selection mistakes where actually quite frequent in this chapter, I'm not sure why exactly considering the experience of an author. $\endgroup$ – ShellRox Feb 4 at 19:17
  • 1
    $\begingroup$ I am pretty sure that rotation matrix in $R^3$ is just the product of 3 individual rotations along x, y and z axis. mathworld.wolfram.com/RotationMatrix.html should give you more information. Since, they are each still individual trigonometric functions, you can do them individually and see that it should still not change the norm $\endgroup$ – Hiten Feb 4 at 19:45
  • 2
    $\begingroup$ What exactly is the book's definition of "a rotation through an angle"? $\endgroup$ – Rahul Feb 4 at 20:59
  • 1
    $\begingroup$ In arbitrary dimension you have to define rotations first. One natural way to so is as "a linear transformation that preserves norms and orientation", but then the claim is trivial. $\endgroup$ – Rahul Feb 4 at 21:26
1
$\begingroup$

If $A$ is a rotation matrix it is orthogonal and $\|A\| = 1$. To keep it consistent with the Euclidean vector norm, I am assuming $\|A\| = \rho(A^T A)$. It is easy to show that $\|A X\| \leq \|A\| \|X\| = \|X\|$, so the proof is complete if we show that $\|AX\| \ge \|X\|$. This last part is easily obtained knowing that $A^{-1} = A^T$.

$\endgroup$
  • $\begingroup$ Thank you for the answer. I was unaware that $||A||=1$. If I may ask, do all orthogonal matrices (rotations, reflections etc.) have $1$ as a norm due to having orthonormal unit vectors as basis? Also, what does $p$ represent in the answer? $\endgroup$ – ShellRox Feb 4 at 21:01
  • 1
    $\begingroup$ Well, it is easier to obtain the result just from the fact that $A A^T = I$. Again, taking this relation gives one inequality and you must prove the equality by finding a specific vector that attains the supremum in the definition of matrix norm. $\rho$ stands for spectral radius and is the matrix norm induced by the Euclidean norm. $\endgroup$ – PierreCarre Feb 4 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.