Mjollnir

Over Specified

Sampled data often provides more equations than there are unknowns. Let's look now at the over-specified problem. We'll use the same original equations but drop out the `z` values.

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

We'll apply the same row and column operations to get at the "inverse". Begin with subtracting one of column `1` from column `2`.

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

Add row `3` to row `2` and subtract 2 of row `3` from row `1`.

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

Now swap rows `1` and `3`.

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

The third row is identically `0`, as expected, so we'll multiply the lower left matrix and the top two rows of the upper right.

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

This is now our `A-1`. We'll confirm this by multiplying it times the `A` matrix.

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

With the `A-1 • A` being the identity matrix, there's not much point in trying to find a null-space. There's not much point in calculating the `(A-1 • A) • A-1` or the `A • (A-1 • A)` because anything times identity or identity times anything is itself. However, it is worth noting that `A • A-1` is not identity.

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

I may come back to this if I can find the time.

For the next part we'll do the same using row and column operations on the under-specified problem.