I Nonempty Subspace: Proving 0u = 0

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Mr Davis 97
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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}##?
 
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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?
 
Mr Davis 97 said:
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.
 

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