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Vector space property

  1. Sep 15, 2009 #1
    I'm considering the problem: Given [tex]c \in \bold{F}[/tex], [tex]v \in V[/tex] where F is a field and V a vector space, show that [tex]cv = 0, v \neq 0 \ \Rightarrow \ c = 0 [/tex]

    I've been wrapping my head around this one for a while now but I can't seem to get it. Proving that if cv = 0 and v [tex]\neq[/tex] 0 implies v = 0 is easy since we can simply multiply by [tex]c^{-1}[/tex] but in vector space, we don't have that kind of inverse for vectors seeing how we only have scalar multiplication.
     
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  3. Sep 15, 2009 #2
    But can you not simply multiply by the inverse of c to get a contradiction - that is v=0?
     
  4. Sep 15, 2009 #3

    HallsofIvy

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    Since c is NOT a vector that doesn't matter!
     
  5. Sep 15, 2009 #4
    When stuck, think about the components. It will get you more stuck or get you an easy proof.
     
  6. Sep 16, 2009 #5
    A "fancy" way to do it if you don't like the coordinate way is to take the alternate definition of a vector space as an embedding of a field into the endomorphism ring of an abelian group. The map for the embedding is just:

    [tex] \theta: F \rightarrow End(A), c \rightarrow cI[/tex]

    Where I is the identity matrix. Then if [tex] v \in A[/tex] what we normally write as [tex]cv[/tex] is just [tex] \theta(c)(v)[/tex] and you can use that to read off all of the vector space properties.

    Then [tex] \theta(c)(v) = 0[/tex] means that [tex] v \in Ker(\theta(c))[/tex] and that should give you the result that you're looking for considering that [tex] \theta(c)[/tex] is invertible by virtue of being in a subfield of End(A).

    EDIT: I should note that you can't use the map that I gave you in the definition because we're conflating the Matn(A) with End(A) but they are isomorphic. I gave it to you so that you would know what the map is. In reality, you would prove that the map induces a ring homomorphism as advertised. This is just a simple exercise. To take it as the definition of a vector space, you assume that the embedding is given.
     
    Last edited: Sep 16, 2009
  7. Oct 3, 2009 #6

    Esd

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    Suppose cv=0, v =/= 0

    then take a new scalar from F (call it a)

    a*(cv)=a*0
    (ac)(v)=0
    ca(v)=0
    c(av)=0

    Then c * (the whole vector space spanned by v) = 0
    Since this is a vector space, c must be 0 (I forget the name of this property).
     
  8. Oct 3, 2009 #7
    the vector spaces are over fields, the components of a vector are elements of the field
     
  9. Oct 6, 2009 #8

    Landau

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    I guess Vee's method is the simplest:

    Assume, by contradiction, that [tex]cv=0[/tex] and [tex]v\neq 0[/tex], but [tex]c\neq 0[/tex]. Then the inverse [tex]c^{-1}[/tex] exists, and multiplying [tex]cv=0[/tex] to the left with it we get [tex]c^{-1}cv=c^{-1}0[/tex], i.e. [tex]v=0[/tex], in contradiction with [tex]v\neq 0[/tex].
     
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