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Additive identity over linear transformation

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  • #1
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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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  • #2
CompuChip
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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.
 
  • #3
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Woops, I meant [itex]\forall\vec{v}\in V,\vec{v}+0=\vec{0+v=\vec{v}}. [/itex] Not the additive inverse.
 
  • #4
CompuChip
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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).
 
  • #5
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Here's a hint: Consider [itex]T(0_v+0_v)[/itex].
 
  • #6
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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]
 
  • #7
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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]
 
  • #8
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[itex]T(0_{v})=0_{w}[/itex], or the additive identity in W?
 
  • #9
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Exactly , it must be the zero of W.
 
  • #10
Fredrik
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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]
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.
 
  • #11
CompuChip
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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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