Linear Tranformation: Find the kernel of T

In summary: If I have a polynomial ##p=a+bx+cx^2+dx^3+ex^4##, which entries in your matrix representation should I use to represent it? Then, what is the kernel of your transformation?For example, in the canonical basis, the polynomial ##p=2+1x+0x^2+3x^3+4x^4## would be represented by the vector ##(2,1,0,3,4)##. Then, you can say your kernel is the space of polynomials for which the vector ##(2,1,0,3,4)## is the null vector.
  • #1
MozAngeles
101
0

Homework Statement


Let T: P4--->P3 be a linear transformation given by T(p)=p'. What is the kernel of T?

Homework Equations

The Attempt at a Solution


T(a0+a1+a2x2+a3x3+a4x4)=a1+2a2x+3a3x2+4a4x3

Ker(T)= { T(p)=0}

so, a1+2a2x+3a3x2+4a4x3=0
then a1=2a2x+3a3x2+4a4x3

Ker(T)= { (-2,1,0,0), (-3,0,1,0), (-4,0,0,1)}

I solved this based off an example from class. But when I checked the dim[Ker(T)]= 3 and the dim[Rng(T)]=4 since my Rng(T)= {1,x,x2,x3} and dim[P4]=5
using general rank nullity theorem I have 7=5, which doesn't make sense. So I'm wondering where I went wrong.

Thank you for your help.
 
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  • #2
MozAngeles said:

Homework Statement


Let T: P4--->P3 be a linear transformation given by T(p)=p'. What is the kernel of T?

Homework Equations

The Attempt at a Solution


T(a0+a1+a2x2+a3x3+a4x4)=a1+2a2x+3a3x2+4a4x3

Ker(T)= { T(p)=0}

so, a1+2a2x+3a3x2+4a4x3=0
then a1=2a2x+3a3x2+4a4x3

Ker(T)= { (-2,1,0,0), (-3,0,1,0), (-4,0,0,1)}

I solved this based off an example from class. But when I checked the dim[Ker(T)]= 3 and the dim[Rng(T)]=4 since my Rng(T)= {1,x,x2,x3} and dim[P4]=5
using general rank nullity theorem I have 7=5, which doesn't make sense. So I'm wondering where I went wrong.

Thank you for your help.
Ker(T) contains all polynomials in P4 whose derivative is 0.

Look at your answer: Ker(T)= { (-2,1,0,0), (-3,0,1,0), (-4,0,0,1)}.
A polynomial in P4 is defined by 5 coefficients. What polynomials in P4 do these elements consisting of 4 coefficients represent?
 
  • #3
Samy_A said:
Ker(T) contains all polynomials in P4 whose derivative is 0.

Look at your answer: Ker(T)= { (-2,1,0,0), (-3,0,1,0), (-4,0,0,1)}.
A polynomial in P4 is defined by 5 coefficients. What polynomials in P4 do these elements consisting of 4 coefficients represent?
I realize where i went wrong and found my kernel to be (0,0,0,0,1) , which then the rest makes sense
 
  • #4
MozAngeles said:
I realize where i went wrong and found my kernel to be (0,0,0,0,1) , which then the rest makes sense
That is correct, sort of.
A good examinator would ask what (0,0,0,0,1) actually represents in P4. In other words, what set of basis vectors are you using? Without that information, (0,0,0,0,1) is meaningless as answer.
Moreover, Ker(T) is not a vector, but a subspace.
 
  • #5
Samy_A said:
That is correct, sort of.
A good examinator would ask what (0,0,0,0,1) actually represents in P4. In other words, what set of basis vectors are you using? Without that information, (0,0,0,0,1) is meaningless as answer.
Moreover, Ker(T) is not a vector, but a subspace.
If my thinking is correct...
{(0,0,0,0,1)} represents the set of vectors that are an element of P4 such that the linear transformation of these vectors, in P4, is a homogeneous equation
 
  • #6
MozAngeles said:
If my thinking is correct...
{(0,0,0,0,1)} represents the set of vectors that are an element of P4 such that the linear transformation of these vectors, in P4, is a homogeneous equation
What linear transformation? How does a homogeneous equation come into play? :oldconfused:

Back to basics: Ker(T) is a subspace of P4.
Your answer should say what polynomials are in that subspace. If you want to represent the elements of P4 in some basis, you have to make clear what basis you are using.
You seem to know what the answer is, but you don't convey it clearly at all.
 
  • #7
Samy_A said:
What linear transformation? How does a homogeneous equation come into play? :oldconfused:

Back to basics: Ker(T) is a subspace of P4.
Your answer should say what polynomials are in that subspace. If you want to represent the elements of P4 in some basis, you have to make clear what basis you are using.
You seem to know what the answer is, but you don't convey it clearly at all.
i was using the definition from by book for kernel space. Being that the initial prob stated it was a linear transformation I used the definition: Ker(T)= {v element of V: T(v)=0) which in words is If T: V--> W is any linear transformation, there is an associated homogeneous linear vector equation T(v)=0. In this case Ker(T)= {(a4,a3,a2,a1,a0):(0,0,0,0,a0}
 
  • #8
MozAngeles said:
i was using the definition from by book for kernel space. Being that the initial prob stated it was a linear transformation I used the definition: Ker(T)= {v element of V: T(v)=0) which in words is If T: V--> W is any linear transformation, there is an associated homogeneous linear vector equation T(v)=0. In this case Ker(T)= {(a4,a3,a2,a1,a0):(0,0,0,0,a0}
What Samy_S was asking was, what polynomials are in the kernel? P4 is a (function) space of polynomials.

Writing an answer of (0, 0, 0, 0, a0) is meaningless if we don't know what the basis is that you're using. In particular, we don't know what polynomials are represented by this vector.
 
  • #9
MozAngeles said:
i was using the definition from by book for kernel space. Being that the initial prob stated it was a linear transformation I used the definition: Ker(T)= {v element of V: T(v)=0) which in words is If T: V--> W is any linear transformation, there is an associated homogeneous linear vector equation T(v)=0. In this case Ker(T)= {(a4,a3,a2,a1,a0):(0,0,0,0,a0}
The helpers basically wanted to know how are the entries in your matrix representation are ordered? When you write ##(a,b,c,d,e)##, are your ordering them according to ##\textrm{coeff}(x^0,x^1,x^2,x^3,x^4)## or ##\textrm{coeff}(x^4,x^3,x^2,x^1,x^0)## or ##\textrm{coeff}(x^3,x^1,x^2,x^4,x^0)## etc?
 

What is a linear transformation?

A linear transformation is a mathematical function that maps elements from one vector space to another. It preserves the properties of linearity, which means that the transformation of a sum of two vectors is equal to the sum of the individual transformations of those vectors.

What is the kernel of a linear transformation?

The kernel of a linear transformation is the set of all vectors that are mapped to the zero vector in the codomain. In other words, it is the set of all input vectors that result in an output of zero.

How do you find the kernel of a linear transformation?

To find the kernel of a linear transformation, you can set up a system of equations using the matrix representation of the transformation and solve for the variables. The solutions to this system of equations will give you the vectors in the kernel.

Why is the kernel of a linear transformation important?

The kernel of a linear transformation is important because it gives us information about the transformation itself. It can tell us about the dimension of the vector space and whether the transformation is invertible or not.

What is the relationship between the kernel and the image of a linear transformation?

The kernel and the image of a linear transformation are complementary subspaces of the domain. The kernel contains all the vectors that are mapped to zero, while the image contains all the vectors that are mapped to non-zero values in the codomain.

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