1. Block LU Decomposition: explained
Keiran O’Haire

2. Introduction Method for performing LU decomposition on large matrices
Requires more calculations
Faster due to memory locality

3. The Matrix
A11
A12
A13
A21
A22
A23
A31
A32
A33

4. Notes
Each of the boxes represent sub matrices of the original matrix
The sub matrices’ size are determined by the block size
Example: sub matrix with block size 4 4 2 1 8

5. Matrix Dimensions b
n-b
A11
A12
A13
A21
A22
A23
A31
A32
A33 b
n-b

6. Step 1
LU Decomposition of left-most column
U11
L11
A12
A13
L21
A22
A23
L31
A32
A33

7. Step 2
Solve the rest of the top row using the topmost L
U11
L11
U12
U13
L21
A22
A23
L31
A32
A33

8. Step 2 Breakdown
Example L
Example U 1 3 2 5

9. Column Version
Use one column at a time from the matrix you wish to solve 1 3 2

10. Column Version Cont’d
The result! 3 -7

11. Column Version Cont’d 1 3 5 Now insert the next column
Repeat until all columns are solved 1 3 5

12. Row Version
Same matrices as before, but this time were doing rows
The version is dependent on the language you are using, such as C or FORTRAN due to array orientation 1 3 2 5

13. Row Version Cont’d
This time, solve each of these (one cell at a time) in a row.
i.e. solve [1 0] = [3]
Clearly, for the first, x1=3, and for the second, x1=1 1 3

14. Row Version Cont’d
These calculated x’s are now used in the next row.
i.e., 3(x1=3) + x2 = 2, x2=-7
Then, 3(x1=1) + x2 = 5, x2=2 1 3 2 5

15. Continued
So now we have calculated this portion. Now…..
U11
L11
U12
U13
L21
A22
A23
L31
A32
A33

16. The Next Step U11 L11 U12 U13 L21 L22 U23 L31 L32 A33 U22
Continue recursively
U11
L11
U12
U13
L21
L22
U23
L31
L32
A33
U22

17. The Completed Decomposition
Eventually you end up with the solution
U11
L11
U12
U13
L21
L22
U23
L31
L32
L33
U22
U33

18. Questions?
Questions?