linear-guest

60 Systems of Linear Equations
What if we stop at a different point in elimination? We could multiply
rows so that the entries in the diagonal are 1 next. Note that the EROs that
do this are diagonal. This gives a slightly different factorization.
Example 30 (LDU factorization building from previous example)
⎛
⎞
2 0 −3 1
M = ⎜ 0 1 2 2
⎟
⎝−4 0 9 2
0 −1 1 −1
∼
⎠ E 3E 2 E 1
E 5
∼
The corresponding elementary matrices are
⎛
⎞ ⎛
2 0 −3 1 1 0 − 3 1⎞
2 2
⎜0 1 2 2
⎟
⎝0 0 3 4⎠ E 4
∼ ⎜0 1 2 2
⎟
⎝0 0 3 4⎠
0 0 0 −3 0 0 0 −3
⎛
1 0 − 3 1⎞
⎛
2 2
1 0 − 3 1⎞
2 2
⎜0 1 2 2
⎟
⎝
4
0 0 1 ⎠ E 6
∼ ⎜0 1 2 2
⎟
⎝
4
3
0 0 1 ⎠ =: U
3
0 0 0 −3 0 0 0 1
⎛
1
⎞ ⎛ ⎞ ⎛
⎞
2
0 0 0
1 0 0 0
1 0 0 0
E 4 = ⎜0 1 0 0
⎟
⎝0 0 1 0⎠ , E 5 = ⎜0 1 0 0
⎟
⎝ 1
0 0
3
0⎠ , E 6 = ⎜0 1 0 0
⎟
⎝0 0 1 0⎠ ,
0 0 0 1
0 0 0 1
0 0 0 − 1 3
⎛ ⎞ ⎛ ⎞ ⎛
⎞
2 0 0 0
1 0 0 0
1 0 0 0
E4 −1
= ⎜0 1 0 0
⎟
⎝0 0 1 0⎠ , E−1 5 = ⎜0 1 0 0
⎟
⎝0 0 3 0⎠ , E−1 6 = ⎜0 1 0 0
⎟
⎝0 0 1 0⎠ .
0 0 0 1
0 0 0 1
0 0 0 −3
The equation U = E 6 E 5 E 4 E 3 E 2 E 1 M can be rearranged as
M = (E −1
1 E−1 2 E−1 3 )(E−1 4 E−1 5 E−1 6 )U.
We calculated the product of the first three factors in the previous example; it was
named L there, and we will reuse that name here. The product of the next three
factors is diagonal and we wil name it D. The last factor we named U (the name means
something different in this example than the last example.) The LDU factorization
of our matrix is
⎛
⎞
⎞ ⎛
⎞ ⎛
2 0 −3 1 1 0 0 0 2 0 0 0 1 0 − 3 1⎞
2 2
⎜ 0 1 2 2⎟
⎜ 0 1 0 0⎟
⎜0 1 0 0⎟
⎜0 1 2 2⎟
⎜
⎝
−4 0 9 2
0 −1 1 −1
⎛
⎟
⎠ = ⎜
⎝
−2 0 1 0
0 −1 1 1
60
⎟ ⎜
⎠ ⎝
0 0 3 0
0 0 0 −3
⎟ ⎜
⎟
⎠ ⎝
4
0 0 1 ⎠ .
3
0 0 0 1