Vector Times Magnitude Same Length As Magnitude Times Vector

Theorem
Let $\mathbf u$ and $\mathbf v$ be two vectors in the vector space $\left({G, +_G, \circ}\right)_K$ over a division ring $\left({K, +_K, \times}\right)$ with subfield $\R$ such that $\R \subseteq Z(K)$ with $Z(K)$ the center of $K$

Let $\left\Vert{ \mathbf u }\right\Vert$ and $\left\Vert{ \mathbf v }\right\Vert$ be the lengths of $\mathbf u$ and $\mathbf v$ respectively.

Then $\left\Vert{ \left({ \left\Vert{ \mathbf v }\right\Vert \circ \mathbf u }\right) }\right\Vert = \left\Vert{ \left({ \left\Vert{ \mathbf u }\right\Vert \circ \mathbf v }\right) }\right\Vert$.

General Proof
Let $\mathbf u$ and $\mathbf v$ be two vectors in the vector space $\left({G, +_G, \circ}\right)_K$.

From the second property of a vector space norm, $\left\Vert{ \left({ \left\Vert{ \mathbf v }\right\Vert \circ \mathbf u }\right) }\right\Vert = \left\vert{ \left({ \left\Vert{ \mathbf v }\right\Vert }\right) }\right\vert_K \times \left\Vert{ \mathbf u }\right\Vert$.

But then since $\left\Vert{ \mathbf v }\right\Vert$ is a non-negative real, and the absolute value function is the norm on the reals, $\left\vert{ \left({ \left\Vert{ \mathbf v }\right\Vert }\right) }\right\vert_K = \left\Vert{ \mathbf v }\right\Vert$.

So then$\left\Vert{ \left({ \left\Vert{ \mathbf v }\right\Vert \circ \mathbf u }\right) }\right\Vert = \left\Vert{ \mathbf v }\right\Vert \times \left\Vert{ \mathbf u }\right\Vert$, and clearly likewise $\left\Vert{ \left({ \left\Vert{ \mathbf u }\right\Vert \circ \mathbf v }\right) }\right\Vert = \left\Vert{ \mathbf v }\right\Vert \times \left\Vert{ \mathbf u }\right\Vert$.

Since $\left\Vert{ \mathbf v }\right\Vert, \left\Vert{ \mathbf u }\right\Vert \in \R$ are in the center of $K$, the result follows.

Proof for Vectors in Euclidean Space
Let $\mathbf u = \left({u_1,u_2,\ldots,u_n}\right)$ and $\mathbf v = \left({v_1,v_2,\ldots,v_n}\right)$ be elements of the real vector space $\R^n$ under the Euclidean norm.

Note that $\left\Vert{ \mathbf v }\right\Vert \circ \mathbf u = \left({ \left\Vert{ \mathbf v }\right\Vert \times u_1, \left\Vert{ \mathbf v }\right\Vert \times u_2 , \ldots , \left\Vert{ \mathbf v }\right\Vert \times u_n }\right)$.

Similarly, $\left\Vert{ \left({ \left\Vert{ \mathbf u }\right\Vert \circ \mathbf v }\right) }\right\Vert = \left\Vert{ \mathbf v }\right\Vert \times \left\Vert{ \mathbf u }\right\Vert$, and since $\times$ is commutative, the result follows.