Using the A‡ previously calculated

On the previous page, row and column operations may be combined to create the inverse. On this page we continue with the original equations and examine the arbitrariness or null space of the original matrix. The original equations are reproduced here (partly because copy/paste is cheap):

Which in short form is:

The matrix A is:

The vector x is:

The vector b is:

And our pseudo-inverse from the previous section:

Solving the problem (finding x)

Multiplying the pseudo-inverse times the b vector produces the x vector.
A‡b = x

Verifying the solution

Confirm this actually works by multiplying the original matrix time this vector.
Ax = b

This is a correct solution, and remember that the first two equations were combined to produce the third, which explains the b1+b2 in the above answer.

Variability in the solution

This is one possibility for the x vector. One of many. How many are there? There's an infinite number of possibilities, but they must all fit a certain pattern. So "what is the pattern", you ask? In this case, instead of solving for a point in 3-space, we get a line. Remember that there are an infinite number of points on a line. If two of the equations were created from a third, the solution would be a plane. If all three equations were unknown, the solution would be 3-space.

Here the z vector is completely arbitrary. If the pseudo-inverse is the unique inverse, the rightmost side of this equation is zero.

This equation gives a handle on the variability of the solution. A proper choice of the z vector will give the same answer as obtained by using the Moore-Penrose method.

Using x = A‡b + (I - A‡A) z

First we'll look at the equation we're solving:

This calculation reduces to this vector:

Plug this general solution back into the original equation to see that all the arbitrary values drop out.

which is:

The arbitrary part of the solution completely dropped out of the above equation and the original b vector is returned.

The above demonstrates the general solution on a square but singular matrix. In the next section we develop the algorithm further by looking at an under-specified problem.