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.)

 1 2 3 1 1 1 2 2 -3 -4 0 2 -1 0 2 • 1 = 0 3 -2

Our inverse (`A-1`) was:

 1 2 1 2 3 2 0 0 3 1 2

If we think about our original equation: `A • x = b`, and the usual solution: `(A-1 • A) • x = A-1 • b`, or just `x = A-1 • b`, 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-1 • b` 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-1 • b + (I - A-1 • A) • z`

Here the `z` vector is completely arbitrary.

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

 1 2 1 2 3 1 2 3 1 2 3 2 2 -3 1 4 0 2 0 0 • -1 0 2 = 0 0 0 3 1 2 0 2 1

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

 1 2 3 1 0 -4 0 2 0 0 0 3 0 -2 0

The next part requires an arbitrary vector `z`.

 1 1 r 2 s 3 t

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

 1 2 3 1 2 3 1 2 3 1 2 2 -3 0 -4 0 0 0 0 2 -1 0 2 • 0 1 0 = 0 0 0 3 0 -2 0

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

 1 2 3 1 1 1 0 -4 0 r -4s 2 0 1 0 • s = s 3 0 -2 0 t -2s

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

 1 2 3 1 1 1 1 2 2 -3 -4s -8s +2s +6s 0 2 -1 0 2 • s = 4s -4s = 0 3 -2s

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

Using this value for the `b` vector:

 1 1 a 2 b

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

 1 2 1 1 1 1 2 3 a -4s 2a +3b -4s 2 0 0 • b + s = s 3 1 2 -2s a +2b -2s

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

 1 2 3 1 1 1 1 2 2 -3 2a +3b -4s 4a +6b -8s +2s -3a -6b +6s a 2 -1 0 2 • s = -2a -3b +4s +2a +4b -4s = b 3 a +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.