Linear Algebra: 4 Fundamental Subspaces

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SUMMARY

The discussion focuses on identifying the bases for the four fundamental subspaces of a matrix A derived from a given LU decomposition. The row space is confirmed to be spanned by the rows of matrix U, while the column space is determined by the columns of matrix L. The nullspace is calculated using the row-reduced echelon form of U, yielding a basis of (0, 1, -2, 1). It is concluded that the absence of a zero row in U indicates there is no left nullspace.

PREREQUISITES
  • Understanding of LU decomposition in linear algebra
  • Familiarity with row-reduced echelon form (RREF)
  • Knowledge of the definitions of row space, column space, nullspace, and left nullspace
  • Proficiency in solving systems of linear equations
NEXT STEPS
  • Study the properties of LU decomposition and its applications in solving linear systems
  • Learn how to compute the row space and column space of a matrix
  • Explore methods for finding the nullspace and left nullspace of matrices
  • Practice problems involving the four fundamental subspaces with various matrix sizes
USEFUL FOR

Students and educators in linear algebra, mathematicians, and anyone seeking to deepen their understanding of matrix theory and its applications in solving linear equations.

ChEJosh
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Homework Statement



Without computing A, find the bases for the 4 fundamental subspaces.

[1 0 0][1 2 3 4]
[6 1 0][0 1 2 3]=A=LU
[9 8 1][0 0 1 2]

Homework Equations



N/A

The Attempt at a Solution



There was an "example" in the book. It just showed the answers.
It was:

[1 0 0][1 3 0 5]
[2 1 0][0 0 1 6]=A
[5 0 1][0 0 0 0]
Where
Row Space: Basis (1,3,0,5) and (0,0,1,6)
Column Space: Basis (1,2,5) and (0,1,0)

In the problem we have to do, I take it that the row space's basis is just the 3 rows of U similar the the example. But, I'm unsure of the column space. Is it the 3 columns of L?

And, also for the nullspace, I put U in its row reduced echelon form, and solved for the nullspace as one would normally do. Is this correct?

[1 2 3 4]
[0 1 2 3] -->
[0 0 1 2]

[1 0 0 0]
[0 1 0 -1]
[0 0 1 2]
So, the nullspace basis is (0,1, -2, 1)

And, since there isn't a zero row, there also isn't a left nullspace, correct?

Sorry if that's confusing. And, thank you in advance!
 
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