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The Jordan canonical form of the matrix
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How to find the Jordan canonical form of the matrix $A$ as a function of the complex parameter $q$:
$$A=beginbmatrix 0 & 3 & q \ 1 & 0 & 0 \ 0 & 1 & 0endbmatrix$$
I can find the characteristic equation where $L$ is an eigenvalue $L^3-3L-q=0$
What should I do next? Looking for roots according to the Cardano formula? Or is there another option?
linear-algebra abstract-algebra jordan-normal-form
edited Jul 17 at 22:47
mechanodroid
22.2k52041
asked Jul 17 at 21:04
Bella
154
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up vote
0
down vote
favorite
How to find the Jordan canonical form of the matrix $A$ as a function of the complex parameter $q$:
$$A=beginbmatrix 0 & 3 & q \ 1 & 0 & 0 \ 0 & 1 & 0endbmatrix$$
I can find the characteristic equation where $L$ is an eigenvalue $L^3-3L-q=0$
What should I do next? Looking for roots according to the Cardano formula? Or is there another option?
linear-algebra abstract-algebra jordan-normal-form
edited Jul 17 at 22:47
mechanodroid
22.2k52041
asked Jul 17 at 21:04
Bella
154
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to find the Jordan canonical form of the matrix $A$ as a function of the complex parameter $q$:
$$A=beginbmatrix 0 & 3 & q \ 1 & 0 & 0 \ 0 & 1 & 0endbmatrix$$
I can find the characteristic equation where $L$ is an eigenvalue $L^3-3L-q=0$
What should I do next? Looking for roots according to the Cardano formula? Or is there another option?
linear-algebra abstract-algebra jordan-normal-form
edited Jul 17 at 22:47
mechanodroid
22.2k52041
asked Jul 17 at 21:04
Bella
154
How to find the Jordan canonical form of the matrix $A$ as a function of the complex parameter $q$:
$$A=beginbmatrix 0 & 3 & q \ 1 & 0 & 0 \ 0 & 1 & 0endbmatrix$$
I can find the characteristic equation where $L$ is an eigenvalue $L^3-3L-q=0$
What should I do next? Looking for roots according to the Cardano formula? Or is there another option?
linear-algebra abstract-algebra jordan-normal-form
edited Jul 17 at 22:47
mechanodroid
22.2k52041
asked Jul 17 at 21:04
Bella
154
edited Jul 17 at 22:47
mechanodroid
22.2k52041
edited Jul 17 at 22:47
mechanodroid
22.2k52041
edited Jul 17 at 22:47
mechanodroid
22.2k52041
22.2k52041
asked Jul 17 at 21:04
Bella
154
asked Jul 17 at 21:04
Bella
154
asked Jul 17 at 21:04
Bella
154
154
add a comment |Â
add a comment |Â
1 Answer
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oldest
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up vote
1
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Two choices, really. For most values of $q$ the characteristic polynomial $L^3 - 3 L - q$ has three distinct roots, so the Jordan form is just diagonal. Not that there is any real chance of writing that all out.
The other choice is when there are repeat eigenvalues. The polynomials over a field make a Euclidean ring, and we can try to find the GCD of the polynomial and its (formal) derivative, that being $3L^2 - 3 .$ This factors as $3(L+1)(L-1).$ The possible roots are $1$ and $-1.$
The two special cases that might have non-diagonal Jordan form are then $q=2,$ which has one Jordan form, and then $q=-2,$ which has a different Jordan form. In both cases you can find all the eigenvalues without difficulty; finish those.
answered Jul 17 at 23:30
Will Jagy
97.2k594196
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Two choices, really. For most values of $q$ the characteristic polynomial $L^3 - 3 L - q$ has three distinct roots, so the Jordan form is just diagonal. Not that there is any real chance of writing that all out.
The other choice is when there are repeat eigenvalues. The polynomials over a field make a Euclidean ring, and we can try to find the GCD of the polynomial and its (formal) derivative, that being $3L^2 - 3 .$ This factors as $3(L+1)(L-1).$ The possible roots are $1$ and $-1.$
The two special cases that might have non-diagonal Jordan form are then $q=2,$ which has one Jordan form, and then $q=-2,$ which has a different Jordan form. In both cases you can find all the eigenvalues without difficulty; finish those.
answered Jul 17 at 23:30
Will Jagy
97.2k594196
add a comment |Â
up vote
1
down vote
accepted
Two choices, really. For most values of $q$ the characteristic polynomial $L^3 - 3 L - q$ has three distinct roots, so the Jordan form is just diagonal. Not that there is any real chance of writing that all out.
The other choice is when there are repeat eigenvalues. The polynomials over a field make a Euclidean ring, and we can try to find the GCD of the polynomial and its (formal) derivative, that being $3L^2 - 3 .$ This factors as $3(L+1)(L-1).$ The possible roots are $1$ and $-1.$
The two special cases that might have non-diagonal Jordan form are then $q=2,$ which has one Jordan form, and then $q=-2,$ which has a different Jordan form. In both cases you can find all the eigenvalues without difficulty; finish those.
answered Jul 17 at 23:30
Will Jagy
97.2k594196
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Two choices, really. For most values of $q$ the characteristic polynomial $L^3 - 3 L - q$ has three distinct roots, so the Jordan form is just diagonal. Not that there is any real chance of writing that all out.
The other choice is when there are repeat eigenvalues. The polynomials over a field make a Euclidean ring, and we can try to find the GCD of the polynomial and its (formal) derivative, that being $3L^2 - 3 .$ This factors as $3(L+1)(L-1).$ The possible roots are $1$ and $-1.$
The two special cases that might have non-diagonal Jordan form are then $q=2,$ which has one Jordan form, and then $q=-2,$ which has a different Jordan form. In both cases you can find all the eigenvalues without difficulty; finish those.
answered Jul 17 at 23:30
Will Jagy
97.2k594196
Two choices, really. For most values of $q$ the characteristic polynomial $L^3 - 3 L - q$ has three distinct roots, so the Jordan form is just diagonal. Not that there is any real chance of writing that all out.
The other choice is when there are repeat eigenvalues. The polynomials over a field make a Euclidean ring, and we can try to find the GCD of the polynomial and its (formal) derivative, that being $3L^2 - 3 .$ This factors as $3(L+1)(L-1).$ The possible roots are $1$ and $-1.$
The two special cases that might have non-diagonal Jordan form are then $q=2,$ which has one Jordan form, and then $q=-2,$ which has a different Jordan form. In both cases you can find all the eigenvalues without difficulty; finish those.
answered Jul 17 at 23:30
Will Jagy
97.2k594196
edited Jul 17 at 23:51
answered Jul 17 at 23:30
Will Jagy
97.2k594196
answered Jul 17 at 23:30
Will Jagy
97.2k594196
answered Jul 17 at 23:30
Will Jagy
97.2k594196
97.2k594196
add a comment |Â
add a comment |Â
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