Solve LU Decomposition for Matrix 0 0 1, 1 0 0, 0 1 0

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Homework Help Overview

The discussion revolves around performing LU decomposition on a specific 3x3 matrix. The original poster presents a matrix that requires decomposition and describes their attempts to manipulate it into upper and lower triangular forms.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to augment the matrix with a zero matrix and apply row operations to achieve the desired triangular forms. They express confusion over the outcome, questioning the decomposability of the matrix.
  • Some participants suggest that the decomposition may be trivial and point out the characteristics of the matrices involved, including the role of permutation matrices.
  • Others raise questions about the properties of the matrices resulting from the decomposition process, particularly regarding the non-zero diagonal elements.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the LU decomposition process. Some guidance has been offered regarding the nature of the matrices involved, but there is no explicit consensus on the original poster's approach or the decomposability of the matrix.

Contextual Notes

Participants note the importance of maintaining non-zero diagonals in the triangular matrices and the potential need for a permutation matrix in computational contexts. The original poster's method and assumptions are under scrutiny, particularly regarding the validity of their row operations.

emira
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Homework Statement


0 0 1
1 0 0
0 1 0 decompose this matrix using LU decomposition.


Homework Equations





The Attempt at a Solution



I took this matrix and augmented it with a zero (3 by 3) matrix. Then I performed the same row operations on both...the row operations with the purpose of making the matrix I was given an upper triangular matrix, and the zero matrix a lower triangular matrix. I ended up just switching rows twice and got the identity matrix for the upper matrix and got the matrix zero for the lower matrix...so no result. Does anyone know any other method I could use to solve this problem?

Thank you,
emira
 
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I believe your original LU decomposition is correct. This is a very trivial decomposition, since it is just a permuted identity matrix.
 
But the product of the identity matrix with matrix zero is matrix zero, thus not giving me the matrix i started with. does that it mean the matrix is not decompos-able?
 
Well, I'm not quite sure how you would have learned how to do LU decomposition, but both your upper and lower triangular matrices should have diagonals non-zero, one of which is usually normalized to be all 1's. Furthermore there is a permutation matrix (especially if you're doing anything computational).

For example, if A is the matrix you're trying to put in LU decomposition, then you find L, U and P such that LUP = A.

Thus let [itex]A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}[/itex]

Then L and U are identity, and P = A (the original matrix).

Edit: Obviously, where L is lower triangular, U is upper triangular, and P is a permuted identity matrix
 

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