Re: Cononical Form

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)
.

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