Proving aX is in U if X is in Rn and U is Subspace of Rn

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Let U be a subspace of Rn

If aX is in U, where a is non zero number and X is in Rn, show that X is in U

THis seems so obvious... but i m not sure how to show this by a proof

aX is in U and aX is in Rn for sure and U is a subspace of Rn.
Is it true that if U is closed under scalar multiplication then X is in U ?

Please advise!
 
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Is it true that if U is closed under scalar multiplication then X is in U ?

Yes. And why is U closed under scalar multiplication?
 
Muzza said:
Yes. And why is U closed under scalar multiplication?


Ohh, me sir! Is it by any chance to do with the definition of what a subspace is?
 
And, by the way, why is it important that a be non-zero? What property do non-zero numbers have that zero does not?
 

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