Vector Times Magnitude Same Length As Magnitude Times Vector

Theorem
Given two vectors $\mathbf u$ and $\mathbf v$ of length $\left\|{\mathbf u}\right\|$ and $\left\|{\mathbf v}\right\|$ respectively, $\left\|{\left({\mathbf u \cdot \left\|{\mathbf v}\right\|}\right)}\right\| = \left\|{\left({\left\|{\mathbf u} \right\| \cdot \mathbf v}\right)}\right\|$.

That is, the first vector times the length of the second vector has the same length as the length of the first vector times the second vector.

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, $\left\Vert{ \left({ \mathbf u \cdot \left\Vert{ \mathbf v }\right\Vert }\right) }\right\Vert = \left\Vert{ \mathbf u }\right\Vert \times \left\vert{ \left({ \left\Vert{ \mathbf v }\right\Vert }\right) }\right\vert_K$.

But then since $\left\Vert{ v }\right\Vert$ 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 $\left\Vert{ \left({ \mathbf u \cdot \left\Vert{ \mathbf v }\right\Vert }\right) }\right\Vert = \left\Vert{ \mathbf u }\right\Vert \times \left\Vert{ \mathbf v }\right\Vert$, and clearly the same holds for $\left\Vert{ \left({ \left\Vert{ \mathbf u }\right\Vert \cdot \mathbf v }\right) }\right\Vert$, so the result follows.

Proof for Vectors in Euclidean Space
Let $\mathbf u = \left({u_1,u_2,\ldots,u_n}\right)$ and $\mathbf v = \left({v_1,v_2,\ldots,v_n}\right)$.

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