jordan decomposition and generalized eigenvectors

Jeremy Watts · science forum Guru Wannabe Joined: 24 Mar 2005 Posts: 239

ok, firstly excuse the length of this post and the fact that it is cross
posted.... (i really didnt know the most appropriate NG to send it to....)

anyway, i am using an algorithm to perform a jordan decomposition taken from
'schaums outlines for matrix operations'. the algorithm states on p.82 that
to form a canonical basis (this being the first step in forming a jordan
decomposition) , then :-

Step 1. Denote the multiplicity of lambda as m , and determine the
smallest positive integer p for which the rank of (A - lambda I )^p equals
n-m , where n denotes the number of rows (and columns in A), lambda denotes
an eigenvalue of A and I is the identity matrix.

Step 2. For each integer k between 1 and p, inclusive, compute the
'eigenvalue rank number Nk' as :-
Nk = rank(A - lambda I)^(k-1) - rank(A - lambdaI)^k
Each Nk is the number of generalized eigenvectors of rank k that will appear
in the canonical basis

Step 3. Determine a generalized eigenvector of rank p, and construct the
chain generated by this vector. Each of these vectors is part of the
canonical basis.

Step 4. Reduce each positive Nk (k = 1,2,...,p) by 1. If all Nk are zero
then stop; the procedure is complete for this particular eigenvalue. If not
then continue to Step 5.

Step 5. Find the highest value of k for which Nk is not zero, and determine
a generalized eigenvector of that rank which is linearly independent of all
previously determined generalized eigenvectors associated with lambda. Form
the chain generated by this vector, and include it in the basis. Return to
Step 4.

Now, the matrix I am using the above procedure on is :-

0 0 1 0 i
0 -9+6i 0 1 0
A = 0 0 8 i 1
0 0 0 8 0
0 2i 0 -9 8

Now the eigenvalues and multiplicities are :-

-9+6i with multiplicity 1
8 with multiplicity 3
0 with multiplicity 1

Starting with -9+6i and going through the procedure then i make the value of
p in step 1 as p = 5. This immediately arouses my suspicions as it looks too
high, as Step 3 not only fails to find a generalized eigenvector of rank 5,
but also even if it existed, the vector plus its chain would be of length 5,
and so fill the entire canonical basis with the vectors generated by just
the first eigenvalue .

By the way I am using the definition of a 'generalized eigenvector' as the
one given in the same book, on the same page in fact as the above procedure,
which is :-

"A vector Xm is a generalized eigenvector of rank m for the square matrix A
and associated eigenvalue lambda if :-

(A - lambda I)^m Xm = 0 but (A - lambda I)^(m-1)Xm =/= 0

So, firstly does anyone agree that a generalized eigenvector of rank 5
cannot exist for the matrix A with the eigenvalue -9+6i , and if so what is
going wrong here generally?

thanks

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