Finding a basis for null(T) and range(T)

  • Thread starter tehme1
  • Start date
  • Tags
    Basis
In summary, the range of T is two-dimensional, with a basis of (1, z2). The null space of T is one-dimensional, with a basis of (z).
  • #1
tehme1
2
0

Homework Statement


Let Y:P3(R) onto P2(R) is defined by T(a0+a1z+a2z2+a3z3)=a1+a2z+a3z2. Find bases for null (T) and range (T). What are their dimensions?


Homework Equations





The Attempt at a Solution


Well, I am assuming the range is the a1+a2z+a3z^2, so it has dimension of 3, and basis (1, z, z^2). As for the null (T) I'm guessing it has dimension 1, since P^3 has dim = 4, and range (T) has dim 3, so 4-3 = 1.
As for the basis of null(T) I'm not sure. I'm guessing it is when a1+a2z+a3z^2 = 0. But I'm not sure how you would solve it or really how you figure these out when they are polynomials and not vectors or matrices. Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
tehme1 said:

Homework Statement


Let Y:P3(R) onto P2(R) is defined by T(a0+a1z+a2z2+a3z3)=a1+a2z2+a3z2. Find bases for null (T) and range (T). What are their dimensions?


Homework Equations





The Attempt at a Solution


Well, I am assuming the range is the a1+a2z+a3z^2, so it has dimension of 3
According to what you wrote, the range is two-dimensional. Was this a typo?
a1+a2z2+a3z2
Every degree-three polynomial gets mapped to a polynomial that consists of a constant + a squared term. There are no terms in x or in x3 in the range.
tehme1 said:
, and basis (1, z, z^2). As for the null (T) I'm guessing it has dimension 1, since P^3 has dim = 4, and range (T) has dim 3, so 4-3 = 1.
As for the basis of null(T) I'm not sure. I'm guessing it is when a1+a2z+a3z^2 = 0. But I'm not sure how you would solve it or really how you figure these out when they are polynomials and not vectors or matrices. Thanks.
 
  • #3
yes, it was typo. I made an edit so now the question is as it should be. Thanks
 
  • #4
tehme1 said:
Well, I am assuming the range is the a1+a2z+a3z^2, so it has dimension of 3, and basis (1, z, z^2). As for the null (T) I'm guessing it has dimension 1, since P^3 has dim = 4, and range (T) has dim 3, so 4-3 = 1.
As for the basis of null(T) I'm not sure. I'm guessing it is when a1+a2z+a3z^2 = 0. But I'm not sure how you would solve it or really how you figure these out when they are polynomials and not vectors or matrices. Thanks.
Start by finding out what T does to the individual functions in a basis for the domain function space.
T(1) = ?
T(x) = ?
T(x2) = ?
T(x3) = ?

That might give you some understanding of what exactly gets mapped to 0 in the range.
 

Related to Finding a basis for null(T) and range(T)

1.

What is the purpose of finding a basis for null(T) and range(T)?

Finding a basis for null(T) and range(T) is important in understanding the properties of a linear transformation T. The null space and range of T provide information about the solutions to the equation T(x) = 0 and the possible outputs of T, respectively.

2.

How do you determine the basis for null(T) and range(T)?

To find a basis for null(T), you can solve the equation T(x) = 0 and express the solutions in terms of free variables. The resulting vectors will form a basis for null(T). To determine a basis for range(T), you can apply T to a set of linearly independent vectors and express the resulting vectors in terms of those original vectors. The resulting vectors will form a basis for range(T).

3.

Can there be multiple bases for null(T) and range(T)?

Yes, there can be multiple bases for null(T) and range(T). The basis for null(T) is not unique and can have multiple sets of vectors that span the null space. Similarly, the basis for range(T) is also not unique and can have multiple sets of vectors that span the range.

4.

What is the relationship between the dimensions of null(T) and range(T)?

The dimensions of null(T) and range(T) are related through the rank-nullity theorem, which states that the dimension of the null space plus the dimension of the range is equal to the dimension of the domain. In other words, dim(null(T)) + dim(range(T)) = dim(domain(T)).

5.

How can finding a basis for null(T) and range(T) be applied in real-world scenarios?

Finding a basis for null(T) and range(T) can be useful in solving systems of linear equations, understanding the behavior of functions, and analyzing data in fields such as physics, engineering, and economics.

Similar threads

Tricky Problem: Prove range T = null ##\phi## when null T' has dim 1
    • Calculus and Beyond Homework Help
Replies
8
Views
865
Null space of dual map of T = annihilator of range T
    • Calculus and Beyond Homework Help
Replies
0
Views
588
Prove that range ##T'## = ##(\text{null}\ T)^0##
    • Calculus and Beyond Homework Help
Replies
1
Views
633
Given surjective ##T:V\to W##, find isomorphism ##T|_U## of U onto W.
    • Calculus and Beyond Homework Help
Replies
1
Views
804
Direct Proof that every zero of p(T) is an eigenvalue of T
    • Calculus and Beyond Homework Help
Replies
24
Views
2K
Linear homogenous system with repeated eigenvalues
    • Calculus and Beyond Homework Help
Replies
2
Views
779
Given subspaces ##U \& W##, show they are equal | Linear Algebra
    • Calculus and Beyond Homework Help
Replies
8
Views
1K
Codomain and Range of Linear Transformation
    • Calculus and Beyond Homework Help
Replies
10
Views
2K
On weakly singular equations and Frobenius' method
    • Calculus and Beyond Homework Help
Replies
7
Views
763
Linear Algebra - Kernel and range of T
    • Calculus and Beyond Homework Help
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top