# Linear subspace proof

1. Aug 17, 2009

### boneill3

1. The problem statement, all variables and given/known data

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 $V$

2. Relevant equations

3. The attempt at a solution

Is it allright 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

2. Aug 17, 2009

### snipez90

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.

3. Aug 17, 2009

### boneill3

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