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Additive identity over linear transformation

  1. Nov 21, 2011 #1
    Given vector spaces V, W over a field, and linear transformation [itex]T:V\rightarrow W[/itex], prove [itex]T(0_{v})=0_{w} [/itex] where 0_v and 0_w are additive identities of V and W.

    I'm trying to use the definition of additive identity. So, [itex]\forall\vec{v}\in V,\vec{v}+0=\vec{v+0=0} [/itex]. Where do I go from here?
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  3. Nov 22, 2011 #2


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    I don't really see why v + 0 = 0... unless v = 0.
    However, you may want to look at v + (-v) = 0.
  4. Nov 22, 2011 #3
    Woops, I meant [itex]\forall\vec{v}\in V,\vec{v}+0=\vec{0+v=\vec{v}}. [/itex] Not the additive inverse.
  5. Nov 22, 2011 #4


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    Yep, I guess you can use that as well.
    You will still need the additive inverse at some point though (if not in V, then at least in W to show that T(v) - T(v) = 0W).
  6. Nov 22, 2011 #5
    Here's a hint: Consider [itex]T(0_v+0_v)[/itex].
  7. Nov 22, 2011 #6
    Would this line of reasoning work?

    [itex]T(0_{v})=T(0_{v}+0_{v})=T(0_{v}+\vec{v}-\vec{v}+0_{v})=T(0_{v}+\vec{v})-T(0_{v}+\vec{v})=T(\vec{v})-T(\vec{v})=\vec{w}-\vec{w}=0_{w}. [/itex]
  8. Nov 22, 2011 #7
    That looks good to me! In fact you can make it even simpler If you want by noting [itex]T(0_v)=T(0_v+0_v)=T(0_v) + T(0_v) [/itex] which is only possible if [itex]T(0_v)=?[/itex]
  9. Nov 22, 2011 #8
    [itex]T(0_{v})=0_{w}[/itex], or the additive identity in W?
  10. Nov 22, 2011 #9
    Exactly , it must be the zero of W.
  11. Nov 22, 2011 #10


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    It's correct, but you're using the following two results without proving them
    • T(-x) is the additive inverse of T(x) for all x.
    • -0=0
    The first one is no more obvious than the statement you're trying to prove.

    If you're going to use these results, there's no need to involve a new vector v. You could just write [tex]T(0)=T(0+0)=T(0+(-0))=T(0)+T(-0)=T(0)-T(0)=0.[/tex] I prefer the method Theorem suggested.
  12. Nov 23, 2011 #11


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    An alternative way to write it down is:
    T(0) = T(0 + 0) = T(0) + T(0) because of linearity.
    Now add the additive inverse of T(0) (whatever it is, but there must be one) to both sides, and you get
    T(0) + (-T(0)) = T(0) + T(0) + (-T(0))
    0 = T(0)
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