Sufficient Condition for Vector Equals Inverse iff Zero

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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, $\mathbf v = \mathbf 0$.

$\blacksquare$