Possible Jordan Canonical Forms Given Minimal Polynomial

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I was supposed to find all possible Jordan canonical forms of a $5times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester. I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2times 2$ while the largest Jordan block with eigenvalue 1 is $1times 1$. Did I miss any matrices or interrupt the minimal polynomial incorrectly?



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &1 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &0 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix




linear-algebra abstract-algebra matrices jordan-normal-form




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  • 4




    You must have a Jordan block associated to the eigenvalue 2 of size 2. Otherwise the minimal polynomial would be $(x-1)(x-2)$.
    – Brandon Carter
    Jan 6 '13 at 0:15






  • 1




    I see. So the last matrix is knocked off. Good.
    – Frank White
    Jan 6 '13 at 0:18














up vote
7
down vote

favorite
5












I was supposed to find all possible Jordan canonical forms of a $5times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester. I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2times 2$ while the largest Jordan block with eigenvalue 1 is $1times 1$. Did I miss any matrices or interrupt the minimal polynomial incorrectly?



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &1 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &0 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix




linear-algebra abstract-algebra matrices jordan-normal-form




share|cite|improve this question

















  • 4




    You must have a Jordan block associated to the eigenvalue 2 of size 2. Otherwise the minimal polynomial would be $(x-1)(x-2)$.
    – Brandon Carter
    Jan 6 '13 at 0:15






  • 1




    I see. So the last matrix is knocked off. Good.
    – Frank White
    Jan 6 '13 at 0:18












up vote
7
down vote

favorite
5









up vote
7
down vote

favorite
5






5





I was supposed to find all possible Jordan canonical forms of a $5times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester. I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2times 2$ while the largest Jordan block with eigenvalue 1 is $1times 1$. Did I miss any matrices or interrupt the minimal polynomial incorrectly?



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &1 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &0 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix




linear-algebra abstract-algebra matrices jordan-normal-form




share|cite|improve this question













I was supposed to find all possible Jordan canonical forms of a $5times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester. I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2times 2$ while the largest Jordan block with eigenvalue 1 is $1times 1$. Did I miss any matrices or interrupt the minimal polynomial incorrectly?



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &1 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &2 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &2 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &1 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix



beginpmatrix
2 &0 &0 &0 &0\
0 &2 &0 &0 &0\
0 &0 &1 &0 &0\
0 &0 &0 &1 &0\
0 &0 &0 &0 &1
endpmatrix





linear-algebra abstract-algebra matrices jordan-normal-form





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edited Apr 30 '14 at 4:40









Martin Sleziak

43.4k6111259




43.4k6111259









asked Jan 6 '13 at 0:11









Frank White

567417




567417







  • 4




    You must have a Jordan block associated to the eigenvalue 2 of size 2. Otherwise the minimal polynomial would be $(x-1)(x-2)$.
    – Brandon Carter
    Jan 6 '13 at 0:15






  • 1




    I see. So the last matrix is knocked off. Good.
    – Frank White
    Jan 6 '13 at 0:18












  • 4




    You must have a Jordan block associated to the eigenvalue 2 of size 2. Otherwise the minimal polynomial would be $(x-1)(x-2)$.
    – Brandon Carter
    Jan 6 '13 at 0:15






  • 1




    I see. So the last matrix is knocked off. Good.
    – Frank White
    Jan 6 '13 at 0:18







4




4




You must have a Jordan block associated to the eigenvalue 2 of size 2. Otherwise the minimal polynomial would be $(x-1)(x-2)$.
– Brandon Carter
Jan 6 '13 at 0:15




You must have a Jordan block associated to the eigenvalue 2 of size 2. Otherwise the minimal polynomial would be $(x-1)(x-2)$.
– Brandon Carter
Jan 6 '13 at 0:15




1




1




I see. So the last matrix is knocked off. Good.
– Frank White
Jan 6 '13 at 0:18




I see. So the last matrix is knocked off. Good.
– Frank White
Jan 6 '13 at 0:18










1 Answer
1






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0
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Yes, you did.



Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1. You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2. Using those rules yields precisely your first four matrices. Your fifth matrix is not correct.



Furthermore, you can permute the blocks. Thus,



  • your first matrix yields 3!/2! = 3 Jordan forms,

  • your second and third matrices yield 4!/2! = 12 forms each, and

  • your four matrix yields 4!/3! = 4 forms

for a total of 3 + 2 · 12 + 4 = 31 forms.






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  • It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease.
    – ancientmathematician
    8 hours ago










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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
0
down vote













Yes, you did.



Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1. You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2. Using those rules yields precisely your first four matrices. Your fifth matrix is not correct.



Furthermore, you can permute the blocks. Thus,



  • your first matrix yields 3!/2! = 3 Jordan forms,

  • your second and third matrices yield 4!/2! = 12 forms each, and

  • your four matrix yields 4!/3! = 4 forms

for a total of 3 + 2 · 12 + 4 = 31 forms.






share|cite|improve this answer





















  • It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease.
    – ancientmathematician
    8 hours ago














up vote
0
down vote













Yes, you did.



Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1. You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2. Using those rules yields precisely your first four matrices. Your fifth matrix is not correct.



Furthermore, you can permute the blocks. Thus,



  • your first matrix yields 3!/2! = 3 Jordan forms,

  • your second and third matrices yield 4!/2! = 12 forms each, and

  • your four matrix yields 4!/3! = 4 forms

for a total of 3 + 2 · 12 + 4 = 31 forms.






share|cite|improve this answer





















  • It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease.
    – ancientmathematician
    8 hours ago












up vote
0
down vote










up vote
0
down vote









Yes, you did.



Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1. You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2. Using those rules yields precisely your first four matrices. Your fifth matrix is not correct.



Furthermore, you can permute the blocks. Thus,



  • your first matrix yields 3!/2! = 3 Jordan forms,

  • your second and third matrices yield 4!/2! = 12 forms each, and

  • your four matrix yields 4!/3! = 4 forms

for a total of 3 + 2 · 12 + 4 = 31 forms.






share|cite|improve this answer













Yes, you did.



Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1. You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2. Using those rules yields precisely your first four matrices. Your fifth matrix is not correct.



Furthermore, you can permute the blocks. Thus,



  • your first matrix yields 3!/2! = 3 Jordan forms,

  • your second and third matrices yield 4!/2! = 12 forms each, and

  • your four matrix yields 4!/3! = 4 forms

for a total of 3 + 2 · 12 + 4 = 31 forms.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered 8 hours ago









Maurice P

1,0571530




1,0571530











  • It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease.
    – ancientmathematician
    8 hours ago
















  • It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease.
    – ancientmathematician
    8 hours ago















It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease.
– ancientmathematician
8 hours ago




It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease.
– ancientmathematician
8 hours ago












 

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