jordan basis problem

**transgalactic** · October 7th 2011, 01:00 PM

find th jordan form of A and find P for which $J=P^{-1}AP$
$A=\left(\begin{array}{ccc}1 & -3 & 3\\-2 & -6 & 13\\-1 & -4 & 8\end{array}\right)$
$|kI-A|=\left|\begin{array}{ccc}k-1 & +3 & -3\\2 & k+6 & -13\\+1 & +4 & k-8\end{array}\right|=(k-1)[(k+6)(k-8)+52]-3[2(k-8)+13]-3[8-k-6]$
$=(k-1)[k^{2}-8k+6k-48+52]-3[2k-16+13]-[2-k]=(k-1)[k^{2}-2k+4]-3[2k-3]-[2-k]=k^{3}-2k^{2}+4k-k^{2}+2k-4-6k+9-2+k=k^{3}-3k^{2}+k+3$

the answer in the book is that the caracteristic polinomial is $(x-1)^3$
i didnt get this answer
why??

then regarding the finding of the basis problem
dont know what they have done here
any guidence
??

**Deveno** · October 7th 2011, 04:03 PM

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:

$\begin{bmatrix}1&1&0\\0&1&1\\0&0&1 \end{bmatrix}$

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...

**transgalactic** · October 8th 2011, 02:52 AM

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
??

**Deveno** · October 8th 2011, 03:53 AM

no, i deduced that it has nilpotency 3 from the fact that there is only 1 Jordan block.

**transgalactic** · October 8th 2011, 04:31 AM

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

**Deveno** · October 8th 2011, 04:48 AM

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.

**transgalactic** · October 8th 2011, 05:02 AM

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?

**Deveno** · October 8th 2011, 05:38 AM

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:

$\begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

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:

$A = PJP^{-1}$ right? so

$A - I = PJP^{-1} - I = PJP^{-1} - PIP^{-1} = P(J-I)P^{-1}$

what is the nilpotency of J-I? just look at it...it's obviously 2! J-I =

$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0 \end{bmatrix}$

(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.

**transgalactic** · October 8th 2011, 06:51 AM

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:

$\begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

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:

$A = PJP^{-1}$ right? so

$A - I = PJP^{-1} - I = PJP^{-1} - PIP^{-1} = P(J-I)P^{-1}$

what is the nilpotency of J-I? just look at it...it's obviously 2! J-I =

$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0 \end{bmatrix}$

(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 block
and its
$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0 \end{bmatrix}$
but its not a proper prof because we have to espablish the minimal polinomial to find the smallest block.

our caracteristic polinomial is $p(t)=(x-1)^3$
and the minimal could be $M(t)=(x-1)^3$ or $M(t)=(x-1)^2$
or $M(t)=(x-1)$

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

?

**Deveno** · October 8th 2011, 06:56 AM

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....

**transgalactic** · October 8th 2011, 10:02 AM

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
?

**Deveno** · October 8th 2011, 10:12 AM

there are different ways of solving linear algebra problems. surely you know that by now.

**transgalactic** · October 8th 2011, 10:19 AM

Originally Posted by Deveno

there are different ways of solving linear algebra problems. surely you know that by now.

ok coming beack to the originak question

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)?

**FernandoRevilla** · October 8th 2011, 10:38 AM

Originally Posted by transgalactic

number of jordan blocks=3-the rank of A

No, the number of Jordan blocks associated to an eigenvalue $\lambda$ of $A\in\mathbb{K}^{n\times n}$ is $n-\textrm{rank}(A-\lambda I)$ . In our case $3-\textrm{rank}(A- I)=3-2=1$ .

**transgalactic** · October 8th 2011, 10:40 AM

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?

Thread: jordan basis problem

LinkBack

Thread Tools

Display

jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Re: jordan basis problem

Similar Math Help Forum Discussions

[SOLVED] Change basis to get Jordan canonical form

Jordan Method and basis

I need help with a Guass Jordan elimination problem

Jordan Basis+Form

Jordan Canonical form problem

Search Tags