Vector Times Magnitude Same Length As Magnitude Times Vector

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Theorem

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

Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be the lengths of $\mathbf u$ and $\mathbf v$ respectively.

Then:

$\norm {\paren {\norm {\mathbf v} \circ \mathbf u} } = \norm {\paren {\norm {\mathbf u} \circ \mathbf v} }$


General Proof

Let $\mathbf u$ and $\mathbf v$ be two vectors in the vector space $\struct {G, +_G, \circ}_K$.

From the second property of a vector space norm:

$\norm {\paren {\norm {\mathbf v} \circ \mathbf u } } = \size {\paren {\norm {\mathbf v} } }_K \times \norm {\mathbf u}$

But then since $\norm {\mathbf v}$ is a non-negative real, and the absolute value function is the norm on the reals:

$\size {\paren {\norm {\mathbf v} } }_K = \norm {\mathbf v}$

So then:

$\norm {\paren {\norm {\mathbf v} \circ \mathbf u} } = \norm {\mathbf v} \times \norm {\mathbf u}$

and clearly likewise:

$\norm {\paren {\norm {\mathbf u} \circ \mathbf v} } = \norm {\mathbf v} \times \norm {\mathbf u}$

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

$\blacksquare$


Proof for Vectors in Euclidean Space

Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$ and $\mathbf v = \tuple {v_1, v_2, \ldots, v_n}$ be elements of the real vector space $\R^n$ under the Euclidean norm.

Note that:

$\norm {\mathbf v} \circ \mathbf u = \tuple {\norm {\mathbf v} \times u_1, \norm {\mathbf v} \times u_2, \ldots, \norm {\mathbf v} \times u_n}$
\(\ds \norm {\paren {\norm {\mathbf v} \circ \mathbf u } }\) \(=\) \(\ds \sqrt {\sum_{i \mathop = 1}^n \paren {\norm {\mathbf v} \times u_i }^2}\)
\(\ds \) \(=\) \(\ds \sqrt {\sum_{i \mathop = 1}^n \norm {\mathbf v}^2 \times \paren {u_i}^2}\)
\(\ds \) \(=\) \(\ds \sqrt {\norm {\mathbf v}^2} \times \sqrt {\sum_{i \mathop = 1}^n u_i^2}\)
\(\ds \) \(=\) \(\ds \norm {\mathbf u} \times \norm {\mathbf v}\)

Similarly:

$\norm {\paren {\norm {\mathbf u} \circ \mathbf v} } = \norm {\mathbf v} \times \norm {\mathbf u}$

and since $\times$ is commutative, the result follows.

$\blacksquare$

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