Re: Cononical Form

What do you mean Top Post?

I understand that every vector would be in the null. So, I tried picking an
arbitrary vector in R2. Say, [1,2]. Using this to generate the cycle and
creating the Q does not yield the desired result (i.e. a=Q*J*(inv(Q))?

"Gerry Myerson" <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:gerry-DA1F07.14504301082005@xxxxxxxxxxxxxxxxxxxxx
> In article <okgHe.23755$Ie1.14720@xxxxxxxxxxxxxxxxxxxxxx>,
> "Sanford Geraci" <sageraci@xxxxxxxxxxx> wrote:
>
>> I did see your approach! However, in the reading of my text it appeared
>> as
>> if there is another way by using the null(T-lamda*I)(x)^n where if the
>> power
>> were n+1, this would be the smallest such power such that we obtain
>> zero(the
>> text didnt show your method). however, when trying the text's method I
>> must
>> be missing something since (t-lambda*I)^2 is the zero matrix. ???????
>> Can
>> you help with that method?
>
> 1. PLEASE don't toppost.
>
> 2. How many approaches do you need?
>
> 3. when you write (T-lamda*I)(x)^n, do you mean (T-lamda*I)^n(x)?
>
> 4. If (t-lambda*I)^2 is the zero matrix then every element of R^2 is in
> the nullspace, so you can pick anything that isn't an eigenvector. Is
> that a problem? How does it compare with what the other approach gives?
>
>> "Gerry Myerson" <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
>> news:gerry-CD9C83.10504801082005@xxxxxxxxxxxxxxxxxxxxx
>> > In article <vRqGe.13124$G71.1146@xxxxxxxxxxxxxxxxxxxxxx>,
>> > "Sanford Geraci" <sageraci@xxxxxxxxxxx> wrote:
>> >
>> >> T is the matrix defined (i.e. for transformations). v1 and v2 are the
>> >> columns of T. How does one find the cononical form? I need
>> >> T=(Q)(J)(inv(Q), where J is the Jordan Cononical form and Q is the
>> >> matrix
>> >> whose columns are created from the cycle of generalized eigenvectors.
>> >> From
>> >> the solutions, I can see that x and (T-lamda*I)(x) are the two
>> >> eigenvectors
>> >> in the cycle. However, I can not see how they got x=(1,0) (note:
>> >> some
>> >> books use the notation tr(1,0) to denote a column vector!)
>> >
>> > Please don't toppost.
>> >
>> > Did you see where I wrote, toward the end of my reply, what it means
>> > for a vector to be a generalized eigenvector?
>> >
>> > Can you see how, if you know the matrix, the eigenvector, and the
>> > eigenvalue, you can use that definition to find the generalized
>> > eigenvector?
>> >
>> > Is there still a problem?
>> >
>> >> "Gerry Myerson" <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
>> >> news:gerry-02A78F.16045429072005@xxxxxxxxxxxxxxxxxxxxx
>> >> > In article <OOhGe.14476$h.1517@xxxxxxxxxxxxxxxxxxxxxx>,
>> >> > "Sanford Geraci" <sageraci@xxxxxxxxxxx> wrote:
>> >> >
>> >> >> Given the matrix: [v1,v2], v1=transpose(1,-1), v2=transpose(1,3),
>> >> >> I
>> >> >> know
>> >> >> that the cycle of generalized eigenvectors is v1=(-1,-1) and
>> >> >> v2=(1,0).
>> >> >> I
>> >> >> know that v1 is (t-lamda*I)(v2). However, how does one find
>> >> >> v2=(1,0)
>> >> >> to
>> >> >> be
>> >> >> a generalized eigenvector??
>> >> >
>> >> > Your notation makes no sense.
>> >> >
>> >> > How can you have v1=transpose(1,-1) and also v1=(-1,-1)?
>> >> > How can you have v2=transpose(1,3) and also v2=(1,0)?
>> >> > How can you have v1 is (t-lambda*I)(v2) when t does not appear
>> >> > in either of your formulas for v1?
>> >> >
>> >> > What do you really mean?
>> >> >
>> >> > Anyway, if A is a matrix and v is a non-zero vector and c is a
>> >> > scalar
>> >> > and Av = cv and w is a non-zero vector and Aw = cw + v then w is
>> >> > a generalized eigenvector.
>> >> >
>> >> > --
>> >> > Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
>> >
>> > --
>> > Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
>
> --
> Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)


.

Relevant Pages

  • Re: eigenvalues of powers of a matrix
    ... The application of your "hint" only shows that if you apply A ... repeatedly to an eigenvector v of A, ... A^r that is a scalar multiple of v. ... is _exactly_ a power r multiple of one of the n eigenvectors of A, ...
    (sci.math.num-analysis)
  • Re: Matrix Diagonalization
    ... The all zeroes is not invalid, it is the 'generilized' eigenvector. ... a generalized eigenvector is something else. ... The diagonalization of A is ... If Lapack does a good job giving you *distinct* roots, ...
    (comp.lang.fortran)
  • Re: Cononical Form
    ... >> books use the notation trto denote a column vector!) ... > Can you see how, if you know the matrix, the eigenvector, and the ... >>> Your notation makes no sense. ... >>> a generalized eigenvector. ...
    (sci.math)
  • Re: Matrix Diagonalization
    ... The all zeroes is not invalid, it is the 'generilized' eigenvector. ... a generalized eigenvector is something else. ... The diagonalization of A is ... form (JCF). ...
    (comp.lang.fortran)