Re: Cononical Form

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?

Thanks for your help.
"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)


.

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