- #1

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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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- #1

- 117

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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?

- #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.

However, you may want to look at v + (-v) = 0.

- #3

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- #4

CompuChip

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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) = 0

- #5

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Here's a hint: Consider [itex]T(0_v+0_v)[/itex].

- #6

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[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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- #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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It's correct, but you're using the following two results without proving them

[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]

- T(-x) is the additive inverse of T(x) for all x.
- -0=0

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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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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