Proving Vector Subspaces with Linear Transformations | Homework Help

Click For Summary
SUMMARY

The discussion centers on proving that the kernel of a linear transformation T: V → W is a vector subspace of V. Key points include demonstrating that the zero vector is in the kernel, and that the kernel is closed under addition and scalar multiplication. Participants emphasized the necessity of showing that if u and v are in ker(T), then u + v is also in ker(T), and if u is in ker(T) and c is a scalar, then c*u must also be in ker(T). These properties stem from the fundamental characteristics of linear transformations.

PREREQUISITES
  • Understanding of vector spaces and their properties
  • Familiarity with linear transformations and their definitions
  • Knowledge of kernel and image of linear mappings
  • Basic concepts of vector addition and scalar multiplication
NEXT STEPS
  • Study the properties of vector spaces in detail
  • Learn about the Rank-Nullity Theorem in linear algebra
  • Explore examples of linear transformations and their kernels
  • Investigate the implications of subspace criteria in vector spaces
USEFUL FOR

Students of linear algebra, mathematics educators, and anyone seeking to deepen their understanding of vector spaces and linear transformations.

boneill3
Messages
126
Reaction score
0

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
 
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
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K