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Linear Algebra

  1. Jan 16, 2006 #1
    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!
     
  2. jcsd
  3. Jan 16, 2006 #2
    Yes. And why is U closed under scalar multiplication?
     
  4. Jan 16, 2006 #3

    matt grime

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    Ohh, me sir! Is it by any chance to do with the definition of what a subspace is?
     
  5. Jan 17, 2006 #4

    HallsofIvy

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    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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