# jordan basis problem

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Oct 9th 2011, 01:10 AM
transgalactic
Re: jordan basis problem
why we look at A-I and not A
?
• Oct 9th 2011, 01:49 AM
Deveno
Re: jordan basis problem
because all of A's eigenvalues are 1?
• Oct 9th 2011, 05:46 AM
transgalactic
Re: jordan basis problem
why did they desided that one of the vectors is e1
?
• Oct 9th 2011, 07:25 AM
Deveno
Re: jordan basis problem
for generalized eigenvectors, you pick the first eigenvector v1 like you normally do, and continue until you run out (in this case, one is all there is).

ker(A - I) has basis {(3,1,1)} now we want to pick our next generalized eigenvector so that:

$\begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}3\\1\\1 \end{bmatrix}$

solve this for (x,y,z), which should give you (0,-2,-1), which is in ker((A-I)^2), but not in ker(A-I).

do it again, and you should get e1 as the 3rd generalized eigenbasis vector.
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12