Show Linear Independence of Set with T:V->V Operator

In summary, the conversation discusses the proof that a set consisting of T(v) and its subsequent m-1 powers is linearly independent when T is a linear operator on a vector space over a field F, and the given conditions are met. The suggestion to apply T to both sides of the equation repeatedly is provided as a hint for the proof.
  • #1
mivanova
7
0
Can you help me with this, or at least give me an idea how to proceed:
Let T:V->V be a linear operator on the vector space over the field F. Let v is in V and let m be a positive integer for which v is not equal to 0, T(v) is not equal to 0, ...,T^(m-1)(v) is not equal to 0, but T^m(v) is equal to 0. Show that {v, T(v), ... , T^(m-1)(v)} is linearly independent set.
Thank you!
 
Physics news on Phys.org
  • #2
Given [tex]a_0 v + a_1 T(v) + \cdots + a_{m-1} T^{m-1}(v) = 0[/tex], apply T to both sides repeatedly and see what you get.
 
Last edited:

1. What is linear independence?

Linear independence is a property of a set of vectors in a vector space. It means that none of the vectors in the set can be written as a linear combination of the other vectors in the set. In other words, the only way to get a sum of zero using the vectors is if all of the coefficients are also zero.

2. How do you show linear independence of a set with a T:V->V operator?

To show linear independence of a set with a T:V->V operator, we need to show that the only solution to the equation T(v) = 0 is when v = 0. This can be done by setting up a system of equations using the vectors in the set and solving for the coefficients. If the only solution is when all the coefficients are zero, then the set is linearly independent.

3. Can a set with a T:V->V operator be linearly independent if it contains only one vector?

No, a set with a T:V->V operator cannot be linearly independent if it contains only one vector. This is because any vector multiplied by a non-zero scalar will still be the same vector, so it can always be written as a linear combination of itself.

4. What is the importance of showing linear independence of a set with a T:V->V operator?

Showcasing linear independence of a set with a T:V->V operator is important because it allows us to determine whether the set forms a basis for the vector space. A basis is a set of linearly independent vectors that span the entire vector space, which is crucial in many applications, such as solving systems of linear equations and representing transformations.

5. Are there any techniques or shortcuts for showing linear independence of a set with a T:V->V operator?

Yes, there are several techniques that can be used to show linear independence of a set with a T:V->V operator. These include using the determinant of the matrix formed by the vectors in the set, checking for linear dependence using the null space of the transformation, and using the rank-nullity theorem. It is important to choose the most efficient method depending on the specific set and operator in question.

Similar threads

  • Calculus and Beyond Homework Help
Replies
0
Views
419
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
534
  • Calculus and Beyond Homework Help
Replies
24
Views
675
  • Calculus and Beyond Homework Help
Replies
1
Views
556
  • Calculus and Beyond Homework Help
Replies
1
Views
423
  • Calculus and Beyond Homework Help
Replies
7
Views
342
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
956
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Back
Top