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If v is a nonzero vector in V,then there is exactly one linear transformation T: V -> W such that T(-v) = -T(v)

I believe this is true, however the solution manual said it was false. I proved by construction given that v1,v2,...,vn are the basis vectors for V, let T1, T2 be linear transformations such that T1 =/= T2. then by homogeneity, I got that T1(-v) = T2(-v), but that contradicts that T1,T2 are not equal therefore it is true.

There is exactly one linear transformation T: V -> W for which T(u+v) = T(u-v) for all vectors u and v in V.

I don't even know where to begin with this so any hints are much appreciated and welcome !

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up vote 2 down vote accepted

For the second question, think about what happens when $u$ is the zero vector.

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you would get T(v) = -T(v) by homogeneity. But isn't that only true if v is the zero vector as well – tamefoxes Nov 12 '12 at 5:55
No, that can happen even if $v$ is not the zero vector. But what does it say about the value of $T(v)$? – Gerry Myerson Nov 12 '12 at 5:57
T(v) has a value of 0? – tamefoxes Nov 12 '12 at 6:00
Yes. And, remember, that's for all $v$. – Gerry Myerson Nov 12 '12 at 6:28

If $v$ is a non-zero vector, then every linear transformation satisfies $$T(-v) = - T(v)$$ by scalar multiplication in the definition of linear transformation.

EDIT: didn't see the second half of the question see Gerry's answer.

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