Mjollnir

Null Space

In the previous section we reduced the original matrix and had an additional 0 column to address. We multiplied the original problem times the third column and the result was zero. This simply means the vector was in the null-space, but there's more than one point in the null-space and we're going to figure out how to get at that here. (We're using 3 dimensions here, and so it's convenient to think of 3-space, but the method works for n-space, but is more difficult to conceptualize.)

12311
122-3-40
2-1021=0
3-2

Our inverse (A-1) was:

12
123
200
312

If we think about our original equation: A • x = b, and the usual solution: (A-1 • A) • x = A-1b, or just x = A-1b, we must remember that the (A-1 • A) was not always the identity matrix, and therefore contained more information that we'd lose if we looked at the x = A-1b only. To get access to this additional information it's necessary to bring across the (A-1 • A) component and combine and reduce this to:

x = A-1b + (I - A-1 • A) • z

Here the z vector is completely arbitrary.

Multiply the "inverse" times the original to get (A-1 • A).

12123123
12322-3140
200-102=000
312021

Subtract this from the identity matrix I to get (I - A-1 • A)

123
10-40
2000
30-20

The next part requires an arbitrary vector z.

1
1r
2s
3t

It's interesting to note that the A matrix times this (I - A-1 • A) matrix is zero.

123123123
122-30-40000
2-102010=000
30-20

And therefore the product (I - A-1 • A) • z will be in the null-space of the original A matrix.

12311
10-40r-4s
2010s=s
30-20t-2s

Let's multiply the original A matrix times this vector:

123111
122-3-4s-8s +2s +6s0
2-102s=4s -4s=0
3-2s

The general solution (xgs) is a combination of the solution using A-1b and the arbitrariness contained in the (I - A-1 • A) • z.

Using this value for the b vector:

1
1a
2b

We can multiply the terms to get the full solution for xgs:

12111
123a-4s2a +3b -4s
200b+s=s
312-2sa +2b -2s

If we multiply the original A matrix times this xgs, we get the b vector returned.

123111
122-32a +3b -4s4a +6b -8s +2s -3a -6b +6sa
2-102s=-2a -3b +4s +2a +4b -4s=b
3a +2b -2s

We put no restrictions on the value of the z vector, and it could be anything. As it turns out (in this example) only the second value (the s) is used. Any value for s is legal because it drops out of the problem.

In this 3-dimensional problem, the "answer" is a line. If we had kept all 3 equations, the "answer" would've been a point. If we had removed two equations from the original set, instead of just the one, the "answer" would've been a plane.

Onward to the demonstration page.