# Sufficient Condition for Vector Equals Inverse iff Zero

Jump to navigation
Jump to search

## 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, $\mathbf v = \mathbf 0$.

$\blacksquare$