# Sufficient Condition for Vector Equals Inverse iff Zero It has been suggested that this article or section be renamed: 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. One may discuss this suggestion on 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

 $\displaystyle \mathbf v$ $=$ $\displaystyle \mathbf 0$ 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.

$\operatorname{Char} \left({\mathbb F}\right) = 0$

so $-2 \cdot 1_{\mathbb F} \ne 0$.

Thus from Vector Product is Zero only if Factor is Zero, $\mathbf v = \mathbf 0$.

$\blacksquare$