How to find Jordan blocks from minimal polynomial
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If the characteristic polynomial is $p=(x-λ)^6$ and the minimal polynomial is $p=(x-λ)^4,$ how do we find all the Jordan blocks?
jordan-normal-form minimal-polynomials
edited Aug 12 at 18:56
Maurice P
1,1601630
asked Apr 18 at 16:42
Jone Will
333
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up vote
1
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If the characteristic polynomial is $p=(x-λ)^6$ and the minimal polynomial is $p=(x-λ)^4,$ how do we find all the Jordan blocks?
jordan-normal-form minimal-polynomials
edited Aug 12 at 18:56
Maurice P
1,1601630
asked Apr 18 at 16:42
Jone Will
333
Hint 1: what are the possibilities, given what you know? Hint 2: what is the rank of $A-lambda I$?
– ancientmathematician
Apr 18 at 17:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If the characteristic polynomial is $p=(x-λ)^6$ and the minimal polynomial is $p=(x-λ)^4,$ how do we find all the Jordan blocks?
jordan-normal-form minimal-polynomials
edited Aug 12 at 18:56
Maurice P
1,1601630
asked Apr 18 at 16:42
Jone Will
333
If the characteristic polynomial is $p=(x-λ)^6$ and the minimal polynomial is $p=(x-λ)^4,$ how do we find all the Jordan blocks?
jordan-normal-form minimal-polynomials
edited Aug 12 at 18:56
Maurice P
1,1601630
asked Apr 18 at 16:42
Jone Will
333
edited Aug 12 at 18:56
Maurice P
1,1601630
edited Aug 12 at 18:56
Maurice P
1,1601630
edited Aug 12 at 18:56
Maurice P
1,1601630
1,1601630
asked Apr 18 at 16:42
Jone Will
333
asked Apr 18 at 16:42
Jone Will
333
asked Apr 18 at 16:42
Jone Will
333
333
Hint 1: what are the possibilities, given what you know? Hint 2: what is the rank of $A-lambda I$?
– ancientmathematician
Apr 18 at 17:01
add a comment |Â
Hint 1: what are the possibilities, given what you know? Hint 2: what is the rank of $A-lambda I$?
– ancientmathematician
Apr 18 at 17:01
Hint 1: what are the possibilities, given what you know? Hint 2: what is the rank of $A-lambda I$?
– ancientmathematician
Apr 18 at 17:01
Hint 1: what are the possibilities, given what you know? Hint 2: what is the rank of $A-lambda I$?
– ancientmathematician
Apr 18 at 17:01
add a comment |Â
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From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks:
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda & 1\
0 & lambda\
endpmatrix mboxor$$
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda
endpmatrix,
beginpmatrix
lambda
endpmatrix.$$
answered Aug 11 at 20:51
Maurice P
1,1601630
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks:
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda & 1\
0 & lambda\
endpmatrix mboxor$$
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda
endpmatrix,
beginpmatrix
lambda
endpmatrix.$$
answered Aug 11 at 20:51
Maurice P
1,1601630
add a comment |Â
up vote
0
down vote
From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks:
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda & 1\
0 & lambda\
endpmatrix mboxor$$
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda
endpmatrix,
beginpmatrix
lambda
endpmatrix.$$
answered Aug 11 at 20:51
Maurice P
1,1601630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks:
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda & 1\
0 & lambda\
endpmatrix mboxor$$
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda
endpmatrix,
beginpmatrix
lambda
endpmatrix.$$
answered Aug 11 at 20:51
Maurice P
1,1601630
From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks:
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda & 1\
0 & lambda\
endpmatrix mboxor$$
$$beginpmatrix
lambda & 1 & 0 & 0\
0 & lambda & 1 & 0\
0 & 0 & lambda & 1\
0 & 0 & 0 & lambda\
endpmatrix,
beginpmatrix
lambda
endpmatrix,
beginpmatrix
lambda
endpmatrix.$$
answered Aug 11 at 20:51
Maurice P
1,1601630
edited Aug 12 at 17:04
answered Aug 11 at 20:51
Maurice P
1,1601630
answered Aug 11 at 20:51
Maurice P
1,1601630
answered Aug 11 at 20:51
Maurice P
1,1601630
1,1601630
add a comment |Â
add a comment |Â
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Hint 1: what are the possibilities, given what you know? Hint 2: what is the rank of $A-lambda I$?
– ancientmathematician
Apr 18 at 17:01