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find th jordan form of A and find P for which
the answer in the book is that the caracteristic polinomial is
i didnt get this answer
why??
then regarding the finding of the basis problem
dont know what they have done here
any guidence
??
http://i51.tinypic.com/qyep34.jpg
check your arithmetic! you dropped a factor of 3 from the term -3[8-k-6] (you should have -3[2-k] at the end, not -[2-k]).
as far as finding P goes, i believe that dim(ker(A-I)) = 1 (prove this!), so the Jordan form of A is:
now A-I has nilpotency 3, right? so we can pick any vector in R^3 for an element in ker((A-I)^3).
it appears your text has chosen v1 = e1, and then chosen
v2= (A-I)v1, which is in ker((A-I)^2), but not in ker((A-I)^3), so...
how did you know that A-I has nilpotency of 3?
its a lot of calulations to do
(A-I)(A-I)=A'
A'A'=0
did you actually calculated these 2 multiplications
??
no, i deduced that it has nilpotency 3 from the fact that there is only 1 Jordan block.
the number of jordan blocks=order of matrices-the rank of A
number of jordan blocks=3-the rank of A
for row reduction i got 3 independant lines
so the rank is 3
so the rank of A=3-3=0
which is wrong?
where is my mistake
you agree that A-I is nilpotent, right? since A is not I, it's nilpotency must be 2 or 3. now we could verify (A-I)^2 is non-zero, or...
we could find the dimension of the eigenspace E1, which is ker(A-I), which will tell us the number of Jordan blocks.
you are checking the rank of the wrong matrix, try checking the rank of A-I.
you must multiply it by itself and get zero and some point in order to establish that its nilpotent
how do you see that its nilpotent just from looking at a matrices?
what is dim(ker(A-I))? if we row-reduce A-I, we find it has rank 2. therefore dim(R^3) = rank(A-I) + dim(ker(A-I)):
3 = 2 + dim(ker(A-I)), so dim(ker(A-I)) = 1.
OR...let's find ker(A-I) explicitly:
right away, we see that x2 = x3.
from either 2 of the remaining equations, we get x1 = 3x2.
so {(3,1,1)} is a basis for A-I. so the Jordan form for A has just one Jordan block, in other words the entire superdiagonal is 1's.
now...you agree that if J is the jordan form for A, that there is some matrix P with:
right? so
what is the nilpotency of J-I? just look at it...it's obviously 2! J-I =
(J-I)^2 will raise the superdiagonal to the upper-right-hand corner, so (J-I)^2 won't be 0, so (A-I)^2 won't be 0.
ok so because the rank of A-I is two its joudan form can have only one blockQuote:
Originally Posted by Deveno
what is dim(ker(A-I))? if we row-reduce A-I, we find it has rank 2. therefore dim(R^3) = rank(A-I) + dim(ker(A-I)):
3 = 2 + dim(ker(A-I)), so dim(ker(A-I)) = 1.
OR...let's find ker(A-I) explicitly:
right away, we see that x2 = x3.
from either 2 of the remaining equations, we get x1 = 3x2.
so {(3,1,1)} is a basis for A-I. so the Jordan form for A has just one Jordan block, in other words the entire superdiagonal is 1's.
now...you agree that if J is the jordan form for A, that there is some matrix P with:
what is the nilpotency of J-I? just look at it...it's obviously 2! J-I =
(J-I)^2 will raise the superdiagonal to the upper-right-hand corner, so (J-I)^2 won't be 0, so (A-I)^2 won't be 0.
and its
but its not a proper prof because we have to espablish the minimal polinomial to find the smallest block.
our caracteristic polinomial is
and the minimal could beor
or
so again i must reprlace x with A and make very long calculations to see if i get the zero matrices in one of them
?
is there another way to find the minimal polinomial
?
the minimal polynomial could only be (x-1)^2 or (x-1)^3.
if (x-1)^2 was the minimal polynomial, (A-I)^2 would be 0, so (J-I)^2 would be 0.
but (J-I)^2 is not 0, it's just not...try it, you'll see (very easy calculation).
suppose x-1 was the minimal polynomial. then A-I = 0 --> A = I. but A is NOT I. so x -1 is not the minimal polynomial. no. isn't. never will be.
that only leaves (x-1)^3. it's our only hope, obi-wan kenobi....
if A-I nilpotent 3 then its largest jordan block is 3
so we dont need minimal polinomial
then why do we need minimal polinomial
in findind jordan form
?
there are different ways of solving linear algebra problems. surely you know that by now.
ok coming beack to the originak questionQuote:
Originally Posted by Deveno
we want to find P:
you said
"now A-I has nilpotency 3, right? so we can pick any vector in R^3 for an element in ker((A-I)^3).
it appears your text has chosen v1 = e1, and then chosen
v2= (A-I)v1, which is in ker((A-I)^2), but not in ker((A-I)^3), so... "
why we can pick any vector in R^3 for ker((A-I)^3)?
No, the number of Jordan blocks associated to an eigenvalueQuote:
Originally Posted by transgalactic
of
is
. In our case
.
so from the jourdan form
if our basis B={v1,v2,v3}
Tv1=1v1+0v2+0v3
Tv2=1v1+1v2+0v3
Tv3=0v1+1v2+1v3
Tv1=Av1
but in the solution they take A-I insteat of A
and they choose e1 e2 e3 in the standart basis
why?