Nonempty Subspace: Proving 0u = 0

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SUMMARY

The discussion centers on proving that for any vector space \( V \), the equation \( 0 \vec{u} = \vec{0} \) holds true for every vector \( \vec{u} \) within \( V \). It is established that this property is inherent to the axioms of vector spaces and does not necessitate the introduction of a specific subspace. The participants emphasize that every subspace is inherently nonempty and adheres to the closure properties of scalar multiplication and vector addition.

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  • Understanding of vector space axioms
  • Familiarity with scalar multiplication in linear algebra
  • Knowledge of subspace properties
  • Basic concepts of vector addition
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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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