PA=LU decomposition (w/ Partial Pivoting)

In summary, the PA=LU decomposition involves three steps: reducing the matrix A to form U using Gaussian Elimination with partial pivoting, forming the matrix L using the "negative of the row reduction multiples," and forming the permutation matrix P based on the row interchanges during the process of pivoting. The process involves selecting rows of the permuted matrix PA, eliminating variables, and retrospectively permuting the columns of L. This process is conceptually straightforward, but in practice, it is done by selecting rows to process and interchanging the rows in A' and L.
  • #1
sid9221
111
0
I'm have a little trouble understanding PA=LU, I have no problems with A=LU but can't seem to figure out the Permutation matrix.

So I have summarised the process I am using let me know where it can be improved.

Step 1: Using Gaussian Elimination with partial pivoting reduce A to form a matrix U.

Step 2: The matrix L is formed by the "negative of the row reduction multiples" eg: R2=R2-(3/4)R3 then -(3/4) is an element in the the matrix L.

Now here is the first problem, as with the A=LU decomposition do these multiples remain fixed in place or do they also permute around depending on row interchanges ?

eg: say R2=R2-(3/4)R3 is the first operation than (-3/4) should go in the position of Row 2, Column 1. But, later if I need to interchange R2 with say R4 (for partial pivoting) will this effect the position of (-3/4) in the matrix L ?

Step 3: Now that you have matrix L and U, form the matrix P with the row interchanges during the process of pivoting.

For example: If during pivoting R1<->R4 and R4<->R3 than

[tex] \begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix} [/tex]

becomes(R1<->R4)

[tex] \begin{bmatrix}
0& 0 & 0 & 1\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
1 & 0 & 0 & 0
\end{bmatrix} [/tex]

and finally (R4<->R3)

[tex]\begin{bmatrix}
0& 0 & 0 & 1\\
0 & 1 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0
\end{bmatrix} [/tex]

The lecture notes I have are extremely complicated and involve L inverse theory which makes my head hurt and I can't find any useful resources online.
 
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  • #2
Conceptually this is straighforward. You decide how to permute all the rows of the matrix somehow, then you do the row interchanges to form P and the matrix product PA, then you factorize PA.

In practice, you don't do it like that. You start by selecting the first row of the permuted matrix PA, then you do the first step of the LU decomposition to "eliminate" the first variable.

That gives you the first column of L, the first row of U, and a matrix A' of size n-1 that contains the "partly processed" numbers from A.

Next, you select the second row to process, based on the numbers in A'. You interchange the rows of A', and you also retrospectively interchange the same rows in the first column of L. You don't do anything to the first row of U.

You then eliminate the second variable, giving the first 2 columns of L, the first 2 rows of U, and a matrix A''of size n-2.

Then you select the third row, and retrospectively permute the first 2 columns of L, and so on.
 

Related to PA=LU decomposition (w/ Partial Pivoting)

1. What is PA=LU decomposition with Partial Pivoting?

PA=LU decomposition with Partial Pivoting is a method used to decompose a square matrix into three matrices: P, L, and U. P is a permutation matrix, L is a lower triangular matrix with ones on the diagonal, and U is an upper triangular matrix. This method is used to solve systems of linear equations and can be more efficient than other methods, especially when the matrix is large and sparse.

2. How does Partial Pivoting work in PA=LU decomposition?

Partial Pivoting is a technique used to avoid division by zero and reduce rounding errors when performing the decomposition. It involves swapping rows in the original matrix to move the largest value in a particular column to the diagonal position. This ensures that the diagonal elements of the U matrix are always the largest values in their respective columns.

3. What is the purpose of the permutation matrix P in PA=LU decomposition?

The permutation matrix P in PA=LU decomposition is used to keep track of the row swaps that occur during Partial Pivoting. This is necessary because the order of the rows in the original matrix affects the resulting L and U matrices. P ensures that the correct rows are swapped and the correct L and U matrices are obtained.

4. How is PA=LU decomposition with Partial Pivoting different from other matrix decomposition methods?

PA=LU decomposition with Partial Pivoting is different from other matrix decomposition methods, such as Gaussian Elimination and Cholesky Decomposition, because it involves both row swaps and column operations. This allows for a more accurate and efficient decomposition, especially for large and sparse matrices.

5. What are the advantages of using PA=LU decomposition with Partial Pivoting?

PA=LU decomposition with Partial Pivoting has several advantages over other matrix decomposition methods. It can handle large and sparse matrices efficiently, it avoids division by zero and reduces rounding errors, and it allows for faster solutions to systems of linear equations. Additionally, the permutation matrix P can be used to efficiently calculate the determinant and inverse of a matrix.

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