- #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.
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.