Does $\Vert \mathbb{U(a)} \times \mathbb{U(b)} \Vert = det(\mathbb{U})\Vert \mathbb{a} \times \mathbb{b} \Vert?$

Question

If $\Vert \mathbb{U(x)}\Vert =\Vert{\mathbb{x}}\Vert$, does $\Vert \mathbb{U(a)} \times \mathbb{U(b)} \Vert = det(\mathbb{U})\Vert \mathbb{a} \times \mathbb{b} \Vert?$

Here $\mathbb{U}:\mathbb{R^n} \to \mathbb{R^n}$ is linear, and $\mathbb{x,a,b} \in \mathbb{R^n}$

My attempt:$\Vert \mathbb{U(a)} \times \mathbb{U(b)} \Vert = \Vert det(\mathbb{U})(\mathbb{U^{-1}})^T(\mathbb{a} \times \mathbb{b}) \Vert $

I think I miss a trivial property of the linear map, can anyone give me a hint?

Any help would be appreciated.

Answer 1

The statement is false, but it is close to a true one. (Also, it's not clear what is meant by the cross product on $\Bbb R^n$.)

Hint So far you have not used the condition $||U({\bf x})|| = ||{\bf x}||$. It is equivalent to $||U({\bf x})||^2 = ||{\bf x}||^2 ,$ which we may rewrite without using the norm as $${\bf x}^\top U^\top U {\bf x} = {\bf x}^\top {\bf x} .$$ Since this holds for all ${\bf x}$, the condition is also equivalent to $$U^\top U = I.$$ Rearranging then gives a simplification for a quantity that appears in the attempt: $$(U^{-1})^\top = U .$$

Remark A (square) matrix that satisfies $U^\top U = I$ is called orthogonal.

Additional hint Applying the transformation identity to $U$ and taking the norm of both sides gives $$||U({\bf a}) \times U({\bf b})|| = ||{\bf a} \times {\bf b}|| .$$ Comparing with the original claim, we see that it holds precisely when $\det U = 1$, that is when $U$ is special orthogonal. In particular, the original claim fails precisely when $\det U = -1$, including in the simple case $U = -I$ mentioned by José in the comments.