Matricies and hypercomplex numbers (again)

[Sam Finch]

a few months ago I posted a fairly strange post about using hypercomplex numbers to represent matricies, I gave up on that afterwards when I realsed I hadn't really made any headway, but I had a new Idea a couple of days ago which works a lot better. The trouble is that there is a special case when an n*n matrix does not have n distinct eigenvectors.

The essential Idea Is this.

In normal complex we base a number system on a number i s.t. i^2 = -1

what I've done is extend this idea to an infinite number of hypercomplex number systems. defining systems Xn based on a number j where j^n = -1 (so each number has n parts in directions 1,j,j^2, j^3, j^4, ... , j^n)

So I worked out the maths for these number systems, going on the idea that for an n*n matrix M there exists a matrix M_bar where M_bar ^ n = -I (I is the Identity matrix)
and where
M = a*I + b*M_bar + c*(M_bar^2) + ... + d*(M_bar^(n-1))

So we can represent M as a number in the system Xn and hence calculate its exponent.


The trouble is M_bar only exists for numbers with unique eigenvectors. For matricies with repeated eigenvectors we have to do something else.

My intuition suggests that it is possible to find an M_bar for a matrix with repeated eigenvectors if we change the rules to say that M_bar^n = 0 (where M_bar^p <> 0 (0<p<n) ) I've defined rules for the corresponding number system but I'm having trouble finding M_bar for matricies larger than 2*2. I can do it fairly easily for M_bar^n = -1, it's just the cases M_bar^n = 0 I'm having trouble with.

Anyone feel like helping?

What comes to my mind is that for the case when you are not getting n unique eigenvectors, then aren't you really dealing with an n*m matrix instead of n*n in the first place?

The repeated eigenvectors are simply not linearly independent?

[Sam Finch]

Sort of. It means one of the rows of the matrix can be expressed as a combination of the other rows (ie the eigenvectors are not linearly independent as you say.) however until you find the eigenvectors it can be quite hard to tell if this is the case.

In matricies used to represent systems of equations a repeated eigenvalue means that there is no unique solution, there are either many solutions or no solutions. However If we need to find the exponential of a matrix chances are it represents a system of differential equations for which there is not usually a unique solution but we are given starting values.

[Sam Finch]

Anybody? To keep hypercomplex stuff out you can answer this.



For an n*n matrix M with at least one repeated eigenvalue, is there an algorithm to find a second matrix M_bar Such that M_bar^n = 0 and there is a polynomial P such that P(M_bar) = M?

[paulw]

I only just remember Eigen values - I last did that just after you were born! Hope somebody else can tho', I am enjoying this after so many years.

Cheers,

P.