Proving Vector Subspaces with Linear Transformations | Homework Help

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boneill3
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Homework Statement



Let V and W be vector spaces over [itex]F[/itex] and [itex]T:V \rightarrow W[/itex] a linear transformation. Prove that [itex]ker(T):=[/itex]{[itex]\epsilon V\mid T()=0_{v}[/itex]} is a vector subspace of [itex]V[/itex]

Homework Equations





The Attempt at a Solution



Is it all right just to state the trivial solution.

ie There exists the vector [itex]0v\epsilon V[/itex] such that
[itex]T(0v) \rightarrow W_{0}[/itex]

therfore the vector [itex]0v\epsilon V[/itex] is also [itex]0v\epsilon T[/itex]

or do I need more Axioms like

There exists the vectors [itex]-v\epsilon V[/itex] and [itex]v\epsilon V[/itex] such that
[itex]T(-v+v) = T(0v) \rightarrow W_{0}[/itex]


to prove that T() is a vector subspace of V
regards
Brendan
 
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You're trying to prove that ker(T) is a subspace, so you have to show that 0 is in ker(T). Moreover, you have to show that if u and v are in ker(T), then so is u + v. Finally, if u is in ker(T) and c is a scalar, then you have to show that c*u is in ker(T). These all follow directly from the basic properties of linear transformations.
 
Thanks for your help.

As a subspace always has the zero vector can I just say that for
{u,v} both elemants of V.
We have
0v = A
0u = B

Ax= 0 and Ay = 0, then A(x + y) = vx + vy = 0 + 0 = 0

Ax = 0 and c is a scalar, then A(cx) = cAx = c0 = 0

{Ax ,Bu} = 0 and c is a scalar, then Acx+Bcy = cAx+cBy = c0 + c0 = 0+0 = 0

Is that allright?



Ax = 0 and c is a scalar, then A(cx) = cAx = c0 = 0