Nonempty Subspace: Proving 0u = 0

[Mr Davis 97]

I have a simple question. Say we have some subspace that is nonempty and closed under scalar multiplication and vector addition. How could we deduce that $0 \vec{u} = \vec{0}$?

 

[S.G. Janssens]

It holds that $0\vec{u} = \vec{0}$ for every $\vec{u} \in V$, where $V$ is any vector space. This follows directly from the defining axioms and does not require the introduction of a subspace.

What did you try yourself to prove it?

 

[member 587159]

I have a simple question. Say we have some subspace that is nonempty and closed under scalar multiplication and vector addition. How could we deduce that $0 \vec{u} = \vec{0}$?

Every subspace is non empty, closed under scalar multiplication and vector addition so no need to say that.

You should show some effort.