What Are the Rank and Nullity of a Linear Transformation?

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Homework Help Overview

The discussion revolves around determining the rank and nullity of a linear transformation T from the space of polynomials of degree at most 5 to itself, specifically focusing on the transformation defined by taking the fourth derivative of a polynomial.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the concepts of rank and nullity, with attempts to identify the basis for the kernel and image of the transformation. Questions arise regarding the nature of the kernel and whether it can be empty or consist of specific polynomials.

Discussion Status

Some participants have provided differing views on the basis of the kernel and image, with ongoing exploration of the implications of the kernel being the zero vector. There is no explicit consensus on the correct basis or the values of rank and nullity.

Contextual Notes

Participants are working under the constraints of the definitions of linear transformations and the properties of polynomial spaces, with some uncertainty about the implications of the kernel's structure on the rank and nullity.

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Homework Statement


find the rank and nullity of the linear transformation T:U -> V and find the basis of the kernel and the image of T


Homework Equations


U=R[x]<=5 V=R[x]<=5 (polynomials of degree at most 5 over R), T(f)=f'''' (4th derivative)


The Attempt at a Solution


Rank = 2
Nullity = 4

basis of kernel = {1,x,x^2,x^3} ?
since a kernel is mapped to V, then the image is the zero vectors? and the basis of the image of T is the empty set?
 
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The basis for Im(T) can be {1, x}.
 
another QUESTION

T : U -> V

and the kernel(T) is the zero vectors,
then what is the basis? it's not the empty set?
 
If it happens, for some linear transformation T, that ker(T) = {0} holds, then yes, the basis for ket(T) is the empty set.
 

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