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 $3×3$, 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 $2×2$ rotation matrix that contains trigonometric functions as columns (before rotation it obviously contains standard basis), but I couldn't find anything similar for $3×3$ matrix.
Thank you!
Best Answer
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 $A^{-1} = A^T$.