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

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Homework Help Overview

The discussion revolves around proving that if a scalar multiple of a vector \( aX \) is in a subspace \( U \) of \( \mathbb{R}^n \), then the vector \( X \) itself must also be in \( U \). The subject area is linear algebra, specifically focusing on properties of vector spaces and subspaces.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of \( U \) being closed under scalar multiplication and question the necessity of the scalar \( a \) being non-zero. There is a discussion about the definition of a subspace and its properties.

Discussion Status

The discussion is active, with participants questioning the definitions and properties related to subspaces. There is an exploration of the implications of closure under scalar multiplication, but no consensus has been reached on the proof itself.

Contextual Notes

Participants are considering the implications of the scalar being non-zero and how that affects the proof. The discussion is framed within the constraints of a homework problem, focusing on theoretical understanding rather than procedural steps.

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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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