Finding Jordan Basis and Form for a Matrix with Zero Eigen Value

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In summary, the conversation discusses finding the Jordan basis and Jordan form of a given matrix. The matrix has a minimal polynomial of X^3, resulting in the Jordan form being the given matrix with the upper right "1" removed. The Jordan basis is found by first finding the cyclic bases, then selecting vectors from Ker(A^2) and not Ker(A). The Jordan basis for the first matrix is {e3-e2, e1, e2}. The conversation also touches on the concept of upper and lower Jordan basis, with the upper one having ones above the diagonal. For the second matrix, the process is similar but requires finding a fourth vector. The conversation also briefly discusses proving that the characteristic polynomial of a matrix divides the
  • #1
MathematicalPhysicist
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i have this matrix:
Code:
011
001
000
and i need to find the matrix jordan basis, and jordan form.
for the jordan bassis i need first to find the eigen value which is zero.
i also found that: Ker(A)={e_1} Ker(A^2)={e_1,e_2} Ker(A^3)=R^3
now Ae1=0, Ae2=e1, Ae3=e1+e2.
now the basis must include e2 and e1, cause we have a transformation from e_2 in Ker(A^2) to e_1 in ker(A), but we don't have a transformation from Ker(A^3) to Ker(A^2), Ae3 should be one of either e2 or e1, can someone help me on this?
thanks in advance.
 
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  • #2
thid matrix has minimal polynomial X^3 hence the jordan form is the given matrix but with the upper right "1" removed.

cant you just replace e3 by e3-e2?
 
Last edited:
  • #3
yes i got this polynomial, my question is how do i find the jordan basis?

as i stated so far e1 and e2 are in this basis but e3 isn't.
there's an algorithm which was shown to me in class, that in order to find the JB, you need first to find the cyclic bases i.e the kernels, and then you find a vector in the basis of R^3 which isn't in Ker(A^2), and then you multiply by A and find the next vector in Ker(A^2) and now you choose a vector which is in Ker(A^2) and not in Ker(A) and do it again.

does it mean that the Jordan basis is {e_2,e_1}?
 
  • #4
re read my post, i added the basis.
 
  • #5
ok, that will do.
and that means that the basis is {e3-e2,e1,e2} right?

ok, i can proceed from here.
thanks.
 
  • #6
well it depends on whetehr you want upper or lower, i gave you upper.didnt i? and yours seems to be neither.

but as you said you can take it from here.
 
  • #7
another question with finding JB:
this time i have this matrix:
Code:
0100
0001
0000
0000
let's denote it A, then A^2 equals:
Code:
0001
0000
0000
0000
and A^3=0.
i got that Ker(A)={e1,e3} Ker(A^2)={e1,e2,e3}
KerA^3=R^4
now Ae4=e2 and Ae2=e1. so iv'e got 3 vectors and i need four, i tried your way by substracting with knowing that Ae3=Ae1=0.
but it doesn't seem to work, i tried with replacing e3 with e2-e3 but it doesn't work as i said already.
if it matters the the char polynomial is x^4 and minimal polynomial is x^3.
 
  • #8
btw, for the first question what is the lower JB?
i know that JB isn't unique, but i haven't heard of upper lower basis.
 
  • #9
well the upper one gives the 1's above the diagonal,...
 
  • #10
isnt Jordan Form is with ones above the diagonal? i guess you can also define it with 1's below, but i haven't learned this part, and i think i wasnt about to find this anyway.
about the second matrix, do you have another simple trick down your sleeve? (im not sure you spell it this way).
 
  • #11
ok i figured out this second matrix.

i have another question, let A be a matrix of nxn, i need to prove that the char polynomial of A divides m(x)^n where m(x) is the minimal polynomial.
well iv'e proven it by stating that m(x) and p(x) have the same linear factors,
m(x)=(x-a1)^l1...(x-a_m)^l_m and p(x)=(x-a1)^k1...(x-a_m)^k_m where k1+...+k_m=n and 1<=l1,..,l_m<=k1,..k_m<=n so by raising m(x) we achieve higher degress of the factors and thus m(x)^n is divisble by p(x).
now iv'e checked in the book and it has a different proof, is my proof valid here?
 

What is a Jordan Basis for a matrix with a zero eigenvalue?

A Jordan Basis for a matrix with a zero eigenvalue is a set of linearly independent vectors that span the nullspace of the matrix. This means that when the matrix is multiplied by any vector in the Jordan Basis, the result will be the zero vector.

Why is it important to find a Jordan Basis for a matrix with a zero eigenvalue?

Finding a Jordan Basis for a matrix with a zero eigenvalue allows us to better understand the structure and properties of the matrix. It also helps us to solve systems of linear equations and perform other calculations involving the matrix.

How can a Jordan Basis be found for a matrix with a zero eigenvalue?

A Jordan Basis can be found by first finding the nullspace of the matrix, which is the set of all vectors that produce the zero vector when multiplied by the matrix. Then, a subset of linearly independent vectors from the nullspace can be chosen to form the Jordan Basis.

What is a Jordan Form for a matrix with a zero eigenvalue?

A Jordan Form for a matrix with a zero eigenvalue is a special type of matrix that is similar to the original matrix, meaning they have the same eigenvalues. However, the Jordan Form has a specific structure with zeros above the main diagonal and ones on the superdiagonal.

Can a matrix have more than one Jordan Basis and Jordan Form with a zero eigenvalue?

Yes, a matrix can have multiple Jordan Bases and Jordan Forms with a zero eigenvalue. This is because the choice of which vectors to include in the Jordan Basis is not unique, and the specific structure of the Jordan Form may vary depending on the basis chosen.

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