I get the only eigenvalue to be zero, dim(Ker(A))=2, dim(Ker(A)^2)=2, then for the cycle tableau, the first column of boxes with have 2 boxes, the second column will have no boxes? I'm confused where to go from there.
#Finding Jordan Form:
lucid tulip
whole sparrowBOT
I get the only eigenvalue to be zero, dim(Ker(A))=2, dim(Ker(A)^2)=2, then for t...
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wispy basin
I get the only eigenvalue to be zero, dim(Ker(A))=2, dim(Ker(A)^2)=2, then for t...
I'd tell you how but doing this type of math is pointless
unless you wanna do something with ai
lucid tulip
i have to for an exam
wispy basin
are you studying this w/ ai purposes?
lucid tulip
no... but isn't this stuff important in other things like calculus and differential equations?
wispy basin
no... but isn't this stuff important in other things like calculus and different...
😭 hell no
literally u barely do any matrice manipulation
fast dirge
Linear Algebra has a lot of applications. It is not "pointless."
wispy basin
Linear Algebra has a lot of applications. It is not "pointless."
for people wanting to be broke sure
fast dirge
It has applications for people wanting to be broke? What does that even mean?
You say you get the dim(Ker(A))=2? I think that dimension should be 1. Then look at (A-0I)^2 = A^2, so find the Ker(A^2).
And then keep going until you have enough generlized eigenvectors.
lucid tulip
oh yeah, i see i wrote down the wrong kernel, thanks
sand quest
@lucid tulip alternatively, you see that the matrix M is nilpotent. You then only need to check that the nilpotency order of M is 4 (because M^3 ≠ 0). Hence, the Jordan normal form of M is J4.
oak riverBOT
@lucid tulip
lucid tulip
+close