Kernel and Linear transformation

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SUMMARY

The discussion focuses on finding the kernel of a linear transformation T: P3 → R, where U is defined as the set of polynomials of degree 3 satisfying the condition 3p(1) = p(0). The basis for U is established as {<1,0,-1,0>,<0,1,-1,0>,<0,0,0,1>}. The kernel is derived from the transformation T, which is represented by the equation T(v) = 0, leading to the conclusion that the kernel consists of linear combinations of the basis vectors resulting in the zero vector.

PREREQUISITES
  • Understanding of polynomial spaces, specifically P3.
  • Knowledge of linear transformations and their properties.
  • Familiarity with kernel and image concepts in linear algebra.
  • Ability to perform vector operations and linear combinations.
NEXT STEPS
  • Study the properties of linear transformations in detail.
  • Learn about the Rank-Nullity Theorem and its applications.
  • Explore polynomial spaces and their bases in linear algebra.
  • Investigate examples of kernels in various linear transformations.
USEFUL FOR

Students and educators in linear algebra, particularly those focusing on polynomial transformations and kernel analysis. This discussion is beneficial for anyone looking to deepen their understanding of linear transformations and their applications in higher mathematics.

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Homework Statement


U = [Polynomial of degree 3 such that 3p(1) = p(0)]

Find the basis of U and find a linear transformation T: P3 ---> R such that U is the kernel of T.

Homework Equations


The Attempt at a Solution



The basis part is easy.
3p(1) = p(0)
3a + 3b + 3c +d = d
c= -b-a

Basis : {<1,0,-1,0>,<0,1,-1,0>,<0,0,0,1>}

The kernel part makes no sense (prof never went over this in class). I know kernel means T(v) = 0 Could the transformation just be R = {a-a, b-b, -(b+a) +(b+a), d-d}
 
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solve:

<1,0,-1,0> * x1 + <0,1,-1,0> * x2 + <0,0,0,1> * x3 = 0
 

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