A^Tx=b given LU factorization for A

[dbkats]

Homework Statement

Suppose you are given the LU factorization for some nxn square matrix A. Assume A is non-singular. This factorization is a result of partial pivoting. Can you use this factorization to solve A^Tx=b for x (given A and b).

Homework Equations

A^T is the transpose of matrix A.
PA = LU is the assumed factorization of A with partial pivoting
Since P is a permutation matrix, P^T=P^-1

The Attempt at a Solution

Haha...I figured it out...

PA=LU
A = (P^T)LU
A^T = (U^T)(L^T)P

A^Tx = (U^T)(L^T)Px = b

Then let Px = y

A^Tx = (U^T)(L^T)y = b

U^T is then lower-triangular, L^T is unit-upper-triangular. Therefore I can solve for y in the usual way, and then figure out what x is based on the permutation matrix P.

 

[mighty2000]

Yes, you can use the LU factorization to solve A^Tx=b for x. As shown in your attempt at a solution, by using the transpose of A and rearranging the equation, you can solve for x in a similar way as you would for the original equation A^Tx=b. This is because the LU factorization allows you to decompose the matrix A into two triangular matrices, which can be easily solved for a given vector b. Additionally, the use of partial pivoting ensures numerical stability and accuracy in the solution.