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Another LU Decomposition Example

Here we will perform an LU decomposition on the matrix
 
1 7 2
M =  −3 −21 4 
1 6 3
following the procedure outlined in Section 7.7.2. So initially we have L 1 =
I 3 and U 1 = M, and hence
   
1 0 0 1 7 2
L 2 =  −3 1 0  U 2 =  0 0 10  .
1 0 1 0 −1 −1

However we now have a problem since 0 · c = 0 for any value of c since we are
working over a field, but we can quickly remedy this by swapping the second and
0
third rows of U 2 to get U and note that we just interchange the corresponding
2
rows all columns left of and including the column we added values to in L 2 to
0 0 0
get L . Yet this gives us a small problem as L U 6= M; in fact it gives us
2 2 2
0
the similar matrix M with the second and third rows swapped. In our original
problem MX = V , we also need to make the corresponding swap on our vector
V to get a V 0 since all of this amounts to changing the order of our two
equations, and note that this clearly does not change the solution. Back to
our example, we have
   
1 0 0 1 7 2
0
0
L =  1 1 0  U =  0 −1 −1  ,
2
2
−3 0 1 0 0 10
0
and note that U is upper triangular. Finally you can easily see that
2
 
1 7 2
0
0
L U =  1 6 = M 0
2 2 3 
−3 −21 4
0
0
0
0
which solves the problem of L U X = M X = V . (We note that as augmented
2
2
0
0
matrices (M |V ) ∼ (M|V ).)
Block LDU Explanation
This video explains how to do a block LDU decomposition. Firstly remember
some key facts about block matrices: It is important that the blocks fit
together properly. For example, if we have matrices
matrix shape
X r × r
Y r × t
Z t × r
W t × t
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