Finding the Kernel and Range of Linear Operators on R3

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SUMMARY

The discussion focuses on determining the kernel and range of the linear operator L(X) = (x1, x1, x1)ᵀ on R³. The kernel is identified as the set of vectors of the form (0, x2, x3), indicating that x1 must be zero. This leads to the conclusion that the kernel is a 2-dimensional subspace, as it can be spanned by the basis vectors {(0, 1, 0), (0, 0, 1)}. The range of the operator is the set of all scalar multiples of the vector (1, 1, 1)ᵀ, which is a 1-dimensional subspace.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically kernel and range of linear operators.
  • Familiarity with vector spaces and subspaces in R³.
  • Knowledge of matrix representation of linear transformations.
  • Ability to identify basis vectors and dimensions of subspaces.
NEXT STEPS
  • Study the properties of linear transformations in R³.
  • Learn about the Rank-Nullity Theorem and its applications.
  • Explore examples of finding the kernel and range for various linear operators.
  • Investigate the concept of basis and dimension in vector spaces.
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Students and educators in linear algebra, mathematicians analyzing linear operators, and anyone seeking to deepen their understanding of vector spaces and subspace dimensions.

Mdhiggenz
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Homework Statement



Determine the kernel/range of each of the following linear operators on R3

L(X)=(x1,x1,x1)T

Homework Equations


The Attempt at a Solution



So first thing I did was create a 3x1 matrix filled with ones.

I equaled it to zero and found x1=0 to be a solution. However I'm not quite sure how they come up with the following answer.

(0,x2,x3). Also why would it be a 2 dimensional subspace? Would it be due to x1 being zero?

Thanks
 
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Mdhiggenz said:

Homework Statement



Determine the kernel/range of each of the following linear operators on R3

L(X)=(x1,x1,x1)T

So first thing I did was create a 3x1 matrix filled with ones.

I equaled it to zero and found x1=0 to be a solution. However I'm not quite sure how they come up with the following answer.

(0,x2,x3). Also why would it be a 2 dimensional subspace? Would it be due to x1 being zero?
x1 = 0 if and only if the vector is of the form (0,x2,x3). The subspace spanned by vectors of this form has dimension 2 because for example {(0,1,0), (0,0, 1)} is a basis.
 
Perfect explanation. Thank you!
 

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