Zero Vector Space Product iff Factor is Zero

Theorem

Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $\left({K, +, \circ}\right)$ be a division ring whose zero is $0$ and whose unity is $1$.

Let $\left({G, +_G, \circ}\right)_K$ be a $K$-vector space.

Let $x \in G, \lambda \in K$.

Then $\lambda \circ x = e \iff \left({\lambda = 0 \lor x = e}\right)$.

Proof 1

A vector space is a module, so all results about modules also apply to vector spaces.

So from Scalar Product with Identity it follows directly that $\lambda = 0 \lor x = e \implies \lambda \circ x = e$.

Next, suppose $\lambda \circ x = e$ but $\lambda \ne 0$.

Then from Scalar Product with Identity:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle e\)	\(=\)	\(\displaystyle \lambda^{-1} \circ e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lambda^{-1} \circ \left({\lambda \circ x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\lambda^{-1} \circ \lambda}\right) \circ x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 \circ x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof 2

The sufficient condition is proved in Vector Scaled by Zero is Zero Vector, and in Zero Vector Scaled is Zero Vector.

The necessary condition is proved in Vector Product is Zero Only If Factor is Zero.

$\blacksquare$

Also see

• Basic Results about Modules
• Basic Results about Unitary Modules