# jordan basis problem

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• Oct 7th 2011, 01:00 PM
transgalactic
jordan basis problem
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 $(x-1)^3$
why??

then regarding the finding of the basis problem
dont know what they have done here
any guidence
??
http://i51.tinypic.com/qyep34.jpg
• Oct 7th 2011, 04:03 PM
Deveno
Re: jordan basis problem
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...
• Oct 8th 2011, 02:52 AM
transgalactic
Re: jordan basis problem
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
??
• Oct 8th 2011, 03:53 AM
Deveno
Re: jordan basis problem
no, i deduced that it has nilpotency 3 from the fact that there is only 1 Jordan block.
• Oct 8th 2011, 04:31 AM
transgalactic
Re: jordan basis problem
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
• Oct 8th 2011, 04:48 AM
Deveno
Re: jordan basis problem
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.
• Oct 8th 2011, 05:02 AM
transgalactic
Re: jordan basis problem
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?
• Oct 8th 2011, 05:38 AM
Deveno
Re: jordan basis problem
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.
• Oct 8th 2011, 06:51 AM
transgalactic
Re: jordan basis problem
Quote:

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

?
• Oct 8th 2011, 06:56 AM
Deveno
Re: jordan basis problem
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....
• Oct 8th 2011, 10:02 AM
transgalactic
Re: jordan basis problem
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
?
• Oct 8th 2011, 10:12 AM
Deveno
Re: jordan basis problem
there are different ways of solving linear algebra problems. surely you know that by now.
• Oct 8th 2011, 10:19 AM
transgalactic
Re: jordan basis problem
Quote:

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)?
• Oct 8th 2011, 10:38 AM
FernandoRevilla
Re: jordan basis problem
Quote:

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 $3-\textrm{rank}(A- I)=3-2=1$ .
• Oct 8th 2011, 10:40 AM
transgalactic
Re: jordan basis problem
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?
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