Sufficient Condition for Vector Equals Inverse iff Zero
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It has been suggested that this page be renamed. In particular: Not sure what to, but current name is obscure, and "Vector Space over Infinite Division Ring implies Vector equals Inverse iff Zero" is just too clumsy. To discuss this page in more detail, feel free to use the talk page. |
Theorem
Let $\left({\mathbf V, +, \circ}\right)_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms.
Let $\mathbb F$ be infinite.
Then:
- $\forall \mathbf v, -\mathbf v \in \mathbf V: \mathbf v = - \mathbf v \iff \mathbf v = \mathbf 0$
Proof
Necessary Condition
\(\ds \mathbf v\) | \(=\) | \(\ds \mathbf 0\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -1_{\mathbb F} \circ \mathbf v\) | \(=\) | \(\ds -1_{\mathbb F} \circ \mathbf 0\) | scaling both sides by the negative of the unity of $\mathbb F$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\mathbf v\) | \(=\) | \(\ds \mathbf 0\) | Vector Inverse is Negative Vector, Zero Vector Scaled is Zero Vector | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf v\) | \(=\) | \(\ds -\mathbf v\) |
$\Box$
Sufficient Condition
Utilizing the vector space axioms:
\(\ds \mathbf v\) | \(=\) | \(\ds -\mathbf v\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf v - \mathbf v\) | \(=\) | \(\ds -\mathbf v - \mathbf v\) | adding $-\mathbf v$ to both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf 0\) | \(=\) | \(\ds \left({-1_{\mathbb F} \circ \mathbf v}\right) + \left({-1_{\mathbb F} \circ \mathbf v}\right)\) | Vector Inverse is Negative Vector | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf 0\) | \(=\) | \(\ds \left({-2 \cdot 1_{\mathbb F} }\right)\circ \mathbf v\) |
$\blacksquare$
By hypothesis, $\mathbb F$ is infinite.
By Characteristic of Division Ring is Zero or Prime:
- $\operatorname{Char} \left({\mathbb F}\right) = 0$
so $-2 \cdot 1_{\mathbb F} \ne 0$.
Thus from Vector Product is Zero only if Factor is Zero, $\mathbf v = \mathbf 0$.
$\blacksquare$
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