Linear Algebra, Theory And Applications, 2012a

356 NORMS FOR FINITE DIMENSIONAL VECTOR SPACES
Multiplying by the inverse of the matrix on the left, 1 this iteration reduces to
⎛
x r+1 ⎞ ⎛
⎞ ⎛ ⎞ ⎛ ⎞
1
1
0 ⎜ x r+1
3
0 0 x r 1
1
2 ⎟
⎝ x r+1 ⎠ = − 1 1
⎜ 4
0
4
0
⎟ ⎜ x r 3
2 ⎟
⎝ 2 1
3
0
x r+1
5
0 ⎠ ⎝ x r ⎠ + 1
⎜ 2 ⎟
⎝ 3 ⎠ . (14.14)
5
3
1
4
0 0 x r 5
4 1
Now iterate this starting with
Thus
Then
⎛
x 2 = − ⎜
⎝
⎛
x 3 = − ⎜
⎝
x 1 ≡
1
0
3
0 0
1 1
4
0
4
0
2 1
0
5
0
5
1
0 0
2
0
1
0
3
0 0
1 1
4
0
4
0
2 1
0
5
0
5
1
0 0
2
0
Continuing this way one finally gets
⎛
x 6 = − ⎜
⎝
1
0
3
0 0
1 1
4
0
4
0
2 1
0
5
0
5
1
0 0
2
0
2
0
⎛
⎜
⎝
⎞ ⎛
⎟ ⎜
⎠ ⎝
0
0
0
0
x 2
0
0
0
0
⎞
⎟
⎠ .
⎞
⎛
⎟
⎠ + ⎜
⎝
⎞ ⎛ { }} ⎞ { ⎛
⎟ ⎜ ⎟
⎠ ⎝ ⎠ + ⎜
⎝
1
3
1
2
3
5
1
x5
1
3
1
2
3
5
1
⎞ ⎛{ }} ⎞{
⎛
. 197
⎟ ⎜ . 351
⎟
⎠ ⎝ . 256 6 ⎠ + ⎜
⎝
. 822
1
3
1
2
3
5
1
⎞
⎛
⎟
⎠ = ⎜
⎝
⎞ ⎛
⎟
⎠ = ⎜
⎝
1
3
1
2
3
5
1
⎞ ⎛
⎟
⎠ = ⎜
⎝
1
3
1
2
3
5
1
. 166
. 26
. 2
. 7
⎞
⎟
⎠
⎞
⎟
⎠
. 216
. 386
. 295
. 871
You can keep going like this. Recall the solution is approximately equal to
⎛ ⎞
. 206
⎜ . 379
⎟
⎝ . 275 ⎠
. 862
so you see that with no care at all and only 6 iterations, an approximate solution has been
obtained which is not too far off from the actual solution.
It is important to realize that a computer would use (14.12) directly. Indeed, writing
the problem in terms of matrices as I have done above destroys every benefit of the method.
However, it makes it a little easier to see what is happening and so this is why I have
presented it in this way.
Definition 14.5.3 The Gauss Seidel method, also called the method of successive corrections
is given as follows. For A =(a ij ) , the iterates for the problem Ax = b are obtained
according to the formula
i∑
j=1
a ij x r+1
j
= −
n∑
j=i+1
⎞
⎟
⎠ .
a ij x r j + b i . (14.15)
1 You certainly would not compute the invese in solving a large system. This is just to show you how the
method works for this simple example. You would use the first description in terms of indices.