The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Matrix Theory

Date: 06/06/2001 at 10:55:32
From: Andre
Subject: Linear Algebra (matrices theory)

I would like to know how to calculate the Jordan canonical form of 
the next matrix, with columns:

[(1 -4 -2 -3 -8),(0 1 -1 -1 -2),(-1 -3 0 -3 -7),(1 2 1 4 5),
   (0 1 1 1 4)].

I've already done it in 3 by 3 matrices, but in this one I'm stuck.
Thank you for your help.

Date: 06/06/2001 at 16:29:34
From: Doctor Rob
Subject: Re: Linear Algebra (matrices theory)

Thanks for writing to Ask Dr. Math, Andre.

Call this matrix M. Then find the characteristic polynomial of M,
det(x*I - M) = f(x). Factor f(x). It turns out to have the form
(x-c)^5. Find the minimal polynomial of M, which is a divisor of this, 
by computing (M-c*I)^j, j = 1, 2, ...  until you get the all-zeroes 
matrix, for j = r. The minimal polynomial is (x-c)^r. Then N = M - c*I 
is a nilpotent matrix, whose minimal polynomial is x^r.

Now the trick is to decompose R^5 into N-cyclic subspaces. There must 
be at least one of dimension r. Picking a random vector for v1, you 
can compute

   v1 = (0,0,1,0,0),
   v1*N = ...
   v1*N^2 = ...
   v1*N^3 = (0,0,0,0,0).

That means that the subspace spanned by v1, v1*N, and v1*N^2 is one of 
the N-cyclic subspaces of dimension 3. Now pick a vectors v2 linearly 
independent of these, and see what the dimensions of the cyclic 
subspaces they generate are. Pick the v2 that gives the largest 
dimension. Continue this until you have the dimensions of cyclic 
subspaces whose direct sum is the whole space,

   R^5 = V1 + V2 + ... + Vk

Then the Jordan canonical form of M is given by k blocks whose sizes
are the dimensions of V1, ..., Vk, each of the form c on the main
diagonal and 1 on the subdiagonal, 0 elsewhere:

       [c  0  0  ...  0]
       [1  c  0  ...  0]
   J = [0  1  c  ...  0]
       [...   ...   ...]
       [0  0  ...  1  c]
These are arranged on the diagonal of a block-diagonal matrix

   [J1  0 ...  0]
   [ 0 J2 ...  0]
   [ ... ... ...]
   [ 0  0 ... Jk]

and they are in decreasing order of size. That is the Jordan Canonical 
Form of the matrix you started with. The basis that gives this form 
for the matrix consists of the vectors

   {N^2*v1, N*v1, v1, ..., vk}

found during this process.

If you have difficulties, feel free to write back.

- Doctor Rob, The Math Forum   
Associated Topics:
College Linear Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.