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

Posted: Tue Jul 18, 2006 6:48 pm Post subject:
jordan decomposition and generalized eigenvectors



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
nm , 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)^(k1)  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)^(m1)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 

Back to top 


Peter Spellucci science forum Guru
Joined: 29 Apr 2005
Posts: 702

Posted: Wed Jul 19, 2006 10:10 am Post subject:
Re: jordan decomposition and generalized eigenvectors



In article <Y1avg.35738$1g.26846@newsfe1win.ntli.net>,
"Jeremy Watts" <jwatts1970@hotmail.com> writes:
Quote:  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
nm , 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)^(k1)  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

this is a misunderstanding:
you have a simple eigenvalue, multiplicity =1
rank deficiency of Alambda*I =1
hence there is a (normal) eigenvector only
hth
peter
be warned: all this is o.k. if you compute anything formally,
roundoff error free. it is absolutely senseless to follow this
approach if you compute numerically. this begins already with
distinguishing multiple eigenvalues and clusters of very nearby eigenvalues
continues with rank determination (under roundoff, many singular matrices
appear as invertible) and the instability of computing the Jordan chain.
if you really must do that numerically, then consult
Golub&van Loan, there a method is described. but also here you will
finally need to know the _strucuture_ of J beforehand in order to get
a save method
Quote:  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)^(m1)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



Back to top 


Helmut Jarausch science forum beginner
Joined: 08 Jul 2005
Posts: 49

Posted: Wed Jul 19, 2006 10:15 am Post subject:
Re: jordan decomposition and generalized eigenvectors



Jeremy Watts wrote:
Quote:  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
nm , 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)^(k1)  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)^(m1)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?

The Jordan form is known to be notoriously instable.
Therefore it shouldn't be used in numerical calculations.
In most cases (e.g. systems of ODEs) the extremely stable 'Schur' form is an
alternative. This gives an orthogonal(!) matrix U and an upper triangular
matrix T s.t. U'*A*U = T (where ' denotes the Hermitian of a matrix =
transposed and complex conjugated).
For your matrix I get (with Scilab (free))
[U,T]=schur(A)
T =
0 1. i 0 0
0 8. 1. 0 i
0 0 8. 2.i  9.
0 0 0  9. + 6.i 1.
0 0 0 0 8.
U =
1. 0 0 0 0
0 0 0 1. 0
0 1. 0 0 0
0 0 0 0 1.
0 0 1. 0 0

Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH  Aachen University
D 52056 Aachen, Germany 

Back to top 


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

Posted: Wed Jul 19, 2006 10:45 am Post subject:
Re: jordan decomposition and generalized eigenvectors



"Helmut Jarausch" <jarausch@igpm.rwthaachen.de> wrote in message
news:44BE063B.50905@igpm.rwthaachen.de...
Quote:  Jeremy Watts wrote:
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
nm , 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)^(k1)  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)^(m1)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?
The Jordan form is known to be notoriously instable.
Therefore it shouldn't be used in numerical calculations.

you can say that again... it wasnt long after posting this that i realised
the problem
was me truncating the eigen value. i was using something like 8.99999...
+ 5.99999i...
to something like 30 decimal places!!! i thought that being this close to
the true
value would make no difference, but in fact it had a catastrophic effect on
the value of
the rank obtained in step 1. the true value should be 1 of course (as
peter pointed out)
but even having a number incredibly close to the real eigenvalue produced an
erroneous value of 5.....
anyway thanks to you and peter for responding.
by the way do either of you know whether there exist any iterative methods
to find a jordan decomposition?
Quote:  In most cases (e.g. systems of ODEs) the extremely stable 'Schur' form is
an
alternative. This gives an orthogonal(!) matrix U and an upper triangular
matrix T s.t. U'*A*U = T (where ' denotes the Hermitian of a matrix =
transposed and complex conjugated).
For your matrix I get (with Scilab (free))
[U,T]=schur(A)
T =
0 1. i 0 0
0 8. 1. 0 i
0 0 8. 2.i  9.
0 0 0  9. + 6.i 1.
0 0 0 0 8.
U =
1. 0 0 0 0
0 0 0 1. 0
0 1. 0 0 0
0 0 0 0 1.
0 0 1. 0 0

Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH  Aachen University
D 52056 Aachen, Germany 


Back to top 


Google


Back to top 



The time now is Mon Aug 21, 2017 8:30 am  All times are GMT

