Sufficient Condition for Vector Equals Inverse iff Zero
There exist infinite fields of characteristic 2
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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
| \(\displaystyle \mathbf v\) | \(=\) | \(\displaystyle \mathbf 0\) | by hypothesis | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle -1_{\mathbb F} \circ \mathbf v\) | \(=\) | \(\displaystyle -1_{\mathbb F} \circ \mathbf 0\) | scaling both sides by the negative of the unity of $\mathbb F$ | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle -\mathbf v\) | \(=\) | \(\displaystyle \mathbf 0\) | Vector Inverse is Negative Vector, Zero Vector Scaled is Zero Vector | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \mathbf v\) | \(=\) | \(\displaystyle -\mathbf v\) |
$\Box$
Sufficient Condition
Utilizing the vector space axioms:
| \(\displaystyle \mathbf v\) | \(=\) | \(\displaystyle -\mathbf v\) | by hypothesis | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \mathbf v - \mathbf v\) | \(=\) | \(\displaystyle -\mathbf v - \mathbf v\) | adding $-\mathbf v$ to both sides | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \mathbf 0\) | \(=\) | \(\displaystyle \left({-1_{\mathbb F} \circ \mathbf v}\right) + \left({-1_{\mathbb F} \circ \mathbf v}\right)\) | Vector Inverse is Negative Vector | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \mathbf 0\) | \(=\) | \(\displaystyle \left({-2 \cdot 1_{\mathbb F} }\right)\circ \mathbf v\) |
$\blacksquare$
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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