Kernal and Range of a Linear Transformation

In summary: So, the basis for ker(L) is {1} and the basis for range(L) is {t^3, t^2}. In summary, the linear transformation L defined by L(p(t)) = t^2 p'(t) has a basis for ker(L) of {1} and a basis for range(L) of {t^3, t^2}. The dimension of ker(L) is 1 and the dimension of range(L) is 2.
  • #1
hkus10
50
0
Let L:p2 >>> p3 be the linear transformation defined by L(p(t)) = t^2 p'(t).
(a) Find a basis for and the dimension of ker(L).
(b) Find a basis for and the dimension of range(L).

The hint that I get is to begin by finding an explicit formula for L by determining
L(at^2 + bt + c).
Does this hint mean let p(t) = at^2 + bt + c?
Then, I find that t^2 p'(t) = 2at^3 + bt^2.
Then, I conclude that the basis for ker(L) = {1}.
Is it right?
Also, how to find range(L)?

Thanks
 
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  • #2
Factorise...

[tex]
L(at^{2}+bt+c)=2at^{3}+bt^{2}=t^{2}(2at+b)
[/tex]

This will be zero when t=0 or t=-b/2a, so...
 
  • #3
hunt_mat said:
Factorise...

[tex]
L(at^{2}+bt+c)=2at^{3}+bt^{2}=t^{2}(2at+b)
[/tex]

This will be zero when t=0 or t=-b/2a, so...

That doesn't have much to do with the problem. hkus10 correctly has ker(L)={1} and having written L(p(t))=2at^3+bt^2 the answer to the dimension and a basis of range(L) should be pretty obvious. Why isn't it hkus10? What's a basis for p3?
 
  • #4
is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?
 
  • #5
hkus10 said:
is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?

Two steps forward, one step backward. Yes, that's a basis for range(L). But now your basis for ker(L) is wrong. I liked your ker(L)={1} a lot better. Why did you put t in? Is t in the kernel?
 
  • #6
hkus10 said:
is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?

I believe you have a workable basis for range(L). However, I think your basis for ker(L) has too many entries.
 
  • #7
Since L(at^2 + bt + c) = 2at^3 + bt^2
No matter what value of t and 1, 2a^3 + bt^2 should always give me 0 vector. So, I have a question why Ker(L) does not have t as a basis?
Another question is dim Ker(L) + dim range(L) = dim (p3) by thm. since the dim range(L) = 2 and dim (p3) = 4, why dim ker(L) not equal to 2?
 
  • #8
hkus10 said:
Since L(at^2 + bt + c) = 2at^3 + bt^2
No matter what value of t and 1, 2a^3 + bt^2 should always give me 0 vector. So, I have a question why Ker(L) does not have t as a basis?
Another question is dim Ker(L) + dim range(L) = dim (p3) by thm. since the dim range(L) = 2 and dim (p3) = 4, why dim ker(L) not equal to 2?

The rank nullity theorem tells you dim ker(L)+dim range(L)=dim(p2). Not dim(p3). dim(p2)=3. t is not in ker(L) because L(t) is not zero. L(t)=t^2. t^2 is not zero.
 
  • #9
Dick said:
The rank nullity theorem tells you dim ker(L)+dim range(L)=dim(p2). Not dim(p3). dim(p2)=3. t is not in ker(L) because L(t) is not zero. L(t)=t^2. t^2 is not zero.

Thanks
 

1. What is a linear transformation?

A linear transformation is a mathematical function that maps one vector space into another. It preserves the structure of vector addition and scalar multiplication, and can be represented by a matrix.

2. What is the kernel of a linear transformation?

The kernel of a linear transformation is the set of all vectors in the domain that transform into the zero vector in the range. In other words, it is the set of all inputs that result in an output of zero.

3. How is the kernel of a linear transformation related to its null space?

The kernel of a linear transformation is the same as its null space. Both refer to the set of vectors that are mapped to zero in the range, and can be represented as the solution space of a homogeneous system of equations.

4. What is the range of a linear transformation?

The range of a linear transformation is the set of all possible outputs that can be produced by applying the transformation to all elements in the domain. It is also known as the image of the linear transformation.

5. How can the kernel and range of a linear transformation be calculated?

The kernel can be calculated by finding the null space of the corresponding matrix representation of the linear transformation. The range can be calculated by applying the transformation to a basis of the domain, and then finding the span of the resulting vectors.

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