Minimal polynomial and possible Jordan forms
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Let $A$ be an $8times 8$ complex matrix with characteristic polynomial $$p_A(x)=(x-1)^4(x+2)^2(x^2+1)$$ and minimal polynomial $$m_A(x)=(x-1)^2(x+2)^2(x^2+1).$$ Determine all possible Jordan canonical forms of $A$. Also, what is the dimension of the eigenspace for $lambda = 1$ for each case?
I haven't learned minimal polynomial yet, so I wanted to check if I am on the right track.
From the characteristic polynomial, the eigenvalues are $i,-i,-2,1$. From the minimal polynomial, the block sizes are $1,1,2$ for eigenvalues $i,-i,-2$ respectively. For $lambda = 1$, the largest block size is 2, so the possible block sizes are $2,2$, which has eigenspace dim of 2 ($J_1$). Or the sizes are $2,1,1$, which has eigenspace dim of 3 ($J_2$). Writing the full Jordan matrix:
$J_1=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&1&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$, $J_2=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&0&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$
linear-algebra jordan-normal-form minimal-polynomials
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up vote
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Let $A$ be an $8times 8$ complex matrix with characteristic polynomial $$p_A(x)=(x-1)^4(x+2)^2(x^2+1)$$ and minimal polynomial $$m_A(x)=(x-1)^2(x+2)^2(x^2+1).$$ Determine all possible Jordan canonical forms of $A$. Also, what is the dimension of the eigenspace for $lambda = 1$ for each case?
I haven't learned minimal polynomial yet, so I wanted to check if I am on the right track.
From the characteristic polynomial, the eigenvalues are $i,-i,-2,1$. From the minimal polynomial, the block sizes are $1,1,2$ for eigenvalues $i,-i,-2$ respectively. For $lambda = 1$, the largest block size is 2, so the possible block sizes are $2,2$, which has eigenspace dim of 2 ($J_1$). Or the sizes are $2,1,1$, which has eigenspace dim of 3 ($J_2$). Writing the full Jordan matrix:
$J_1=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&1&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$, $J_2=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&0&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$
linear-algebra jordan-normal-form minimal-polynomials
2
Looks to me like you did everything right!
â Christian
Jul 19 '16 at 2:37
@Christian yay! thank you
â Julie
Jul 19 '16 at 2:52
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up vote
3
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up vote
3
down vote
favorite
Let $A$ be an $8times 8$ complex matrix with characteristic polynomial $$p_A(x)=(x-1)^4(x+2)^2(x^2+1)$$ and minimal polynomial $$m_A(x)=(x-1)^2(x+2)^2(x^2+1).$$ Determine all possible Jordan canonical forms of $A$. Also, what is the dimension of the eigenspace for $lambda = 1$ for each case?
I haven't learned minimal polynomial yet, so I wanted to check if I am on the right track.
From the characteristic polynomial, the eigenvalues are $i,-i,-2,1$. From the minimal polynomial, the block sizes are $1,1,2$ for eigenvalues $i,-i,-2$ respectively. For $lambda = 1$, the largest block size is 2, so the possible block sizes are $2,2$, which has eigenspace dim of 2 ($J_1$). Or the sizes are $2,1,1$, which has eigenspace dim of 3 ($J_2$). Writing the full Jordan matrix:
$J_1=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&1&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$, $J_2=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&0&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$
linear-algebra jordan-normal-form minimal-polynomials
Let $A$ be an $8times 8$ complex matrix with characteristic polynomial $$p_A(x)=(x-1)^4(x+2)^2(x^2+1)$$ and minimal polynomial $$m_A(x)=(x-1)^2(x+2)^2(x^2+1).$$ Determine all possible Jordan canonical forms of $A$. Also, what is the dimension of the eigenspace for $lambda = 1$ for each case?
I haven't learned minimal polynomial yet, so I wanted to check if I am on the right track.
From the characteristic polynomial, the eigenvalues are $i,-i,-2,1$. From the minimal polynomial, the block sizes are $1,1,2$ for eigenvalues $i,-i,-2$ respectively. For $lambda = 1$, the largest block size is 2, so the possible block sizes are $2,2$, which has eigenspace dim of 2 ($J_1$). Or the sizes are $2,1,1$, which has eigenspace dim of 3 ($J_2$). Writing the full Jordan matrix:
$J_1=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&1&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$, $J_2=beginpmatrix1&1&0&0&0&0&0&0\0&1&0&0&0&0&0&0\0&0&1&0&0&0&0&0\0&0&0&1&0&0&0&0\0&0&0&0&-2&1&0&0\0&0&0&0&0&-2&0&0\0&0&0&0&0&0&i&0\0&0&0&0&0&0&0&-iendpmatrix$
linear-algebra jordan-normal-form minimal-polynomials
asked Jul 19 '16 at 2:30
Julie
21916
21916
2
Looks to me like you did everything right!
â Christian
Jul 19 '16 at 2:37
@Christian yay! thank you
â Julie
Jul 19 '16 at 2:52
add a comment |Â
2
Looks to me like you did everything right!
â Christian
Jul 19 '16 at 2:37
@Christian yay! thank you
â Julie
Jul 19 '16 at 2:52
2
2
Looks to me like you did everything right!
â Christian
Jul 19 '16 at 2:37
Looks to me like you did everything right!
â Christian
Jul 19 '16 at 2:37
@Christian yay! thank you
â Julie
Jul 19 '16 at 2:52
@Christian yay! thank you
â Julie
Jul 19 '16 at 2:52
add a comment |Â
1 Answer
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You identified the blocks and eigenspace dimensions correctly. The blocks for a given form, however, can be permuted. Thus, your first matrix represents 5!/2! = 60 forms, and your second matrix, 6!/2! = 360 forms for a total of 420 forms.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You identified the blocks and eigenspace dimensions correctly. The blocks for a given form, however, can be permuted. Thus, your first matrix represents 5!/2! = 60 forms, and your second matrix, 6!/2! = 360 forms for a total of 420 forms.
add a comment |Â
up vote
0
down vote
You identified the blocks and eigenspace dimensions correctly. The blocks for a given form, however, can be permuted. Thus, your first matrix represents 5!/2! = 60 forms, and your second matrix, 6!/2! = 360 forms for a total of 420 forms.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You identified the blocks and eigenspace dimensions correctly. The blocks for a given form, however, can be permuted. Thus, your first matrix represents 5!/2! = 60 forms, and your second matrix, 6!/2! = 360 forms for a total of 420 forms.
You identified the blocks and eigenspace dimensions correctly. The blocks for a given form, however, can be permuted. Thus, your first matrix represents 5!/2! = 60 forms, and your second matrix, 6!/2! = 360 forms for a total of 420 forms.
answered Aug 8 at 17:10
Maurice P
1,1501630
1,1501630
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2
Looks to me like you did everything right!
â Christian
Jul 19 '16 at 2:37
@Christian yay! thank you
â Julie
Jul 19 '16 at 2:52