Proving Vector Subspaces with Linear Transformations | Homework Help

boneill3
Messages
126
Reaction score
0

Homework Statement



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

Homework Equations





The Attempt at a Solution



Is it all right just to state the trivial solution.

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

therfore the vector 0v\epsilon V is also 0v\epsilon T

or do I need more Axioms like

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


to prove that T() is a vector subspace of V
regards
Brendan
 
Physics news on Phys.org
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top