How do I complete the steps of finding the Jordan of this $5times 5$ matrix (with Octave)? The 2019 Stack Overflow Developer Survey Results Are InDetermining the Jordan Canonical Form $18times 18$ matrixFinding Jordan Normal Form of 4x4 MatrixJordan form of the matrix $left(beginsmallmatrix 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 1 & 1 endsmallmatrixright)$Jordan Normal Form of $A_7 times 7$ with a singal eigenvalue $lambda$Jordan normal form which depends on parametercompute the Jordan matrix $ mathcalM(T) $Determine the JCF from $operatornamerank(A-lambda I)^j$Find the Jordan forms of a matrix from just the ranks of its eigenspacesJordan Normal Form: Two times the same basis vector?!How do I find the bases of the Jordan Canonical Form of $C$?
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How do I complete the steps of finding the Jordan of this $5times 5$ matrix (with Octave)?
The 2019 Stack Overflow Developer Survey Results Are InDetermining the Jordan Canonical Form $18times 18$ matrixFinding Jordan Normal Form of 4x4 MatrixJordan form of the matrix $left(beginsmallmatrix 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 1 & 1 endsmallmatrixright)$Jordan Normal Form of $A_7 times 7$ with a singal eigenvalue $lambda$Jordan normal form which depends on parametercompute the Jordan matrix $ mathcalM(T) $Determine the JCF from $operatornamerank(A-lambda I)^j$Find the Jordan forms of a matrix from just the ranks of its eigenspacesJordan Normal Form: Two times the same basis vector?!How do I find the bases of the Jordan Canonical Form of $C$?
$begingroup$
I know how to begin the procedure but I don't know how to finish it. Let's start with an example (sorry for it being so unwieldy).
Let
$$A =beginpmatrix
177& 548& 271& -548& -356\ 19& 63& 14& -79& -23\ 8& 24& 17& -20& -20\ 42& 132& 55& -141& -76\ 56& 176& 80& -184& -105endpmatrix$$
Find the Jordan canonical form of A and the change of basis matrix.
STEP 1. Find the characteristic polynomial.
You can do this step using the following command
[V, lam] = eig(A)
which produces for the variable lam the eigenvalues with repeats, allowing one to easily deduce the following for the characteristic polynomial.
$chi_A(lambda) = (lambda - 3)^4(lambda + 1)$.
STEP 2. Find the geometric multiplicity for $lambda = 3$.
To do this we must find the nullity, which is equal to $5 - textrank(A - 3I)$ by the rank-nullity theorem. Commanding 5 - rank(A - 3 * eye(5)) gives $2$.
STEP 3. Find the geometric multiplicity for $lambda = -1$.
To do this we must find the nullity, which is equal to $5 - textrank(A + I)$ by the rank-nullity theorem. Commanding 5 - rank(A + eye(5)) gives $1$.
STEP 4. Make a table and try to guess the Jordan form.
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
3 & 4 & 2 \
-1 & 1 & 1
endarray
$$
However in this case we cannot guess, because are two Jordan blocks and their dimensions sum to 4. There are two ways to split 4 into two integers and we don't know which one is it.
How do I proceed? If you could, I would greatly appreciate it if you use the step structure I had been using throughout this post. Ideally with Octave commands.
linear-algebra jordan-normal-form octave
edited Mar 31 at 0:07
idriskameni
751321
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
$endgroup$
|
show 2 more comments
$begingroup$
I know how to begin the procedure but I don't know how to finish it. Let's start with an example (sorry for it being so unwieldy).
Let
$$A =beginpmatrix
177& 548& 271& -548& -356\ 19& 63& 14& -79& -23\ 8& 24& 17& -20& -20\ 42& 132& 55& -141& -76\ 56& 176& 80& -184& -105endpmatrix$$
Find the Jordan canonical form of A and the change of basis matrix.
STEP 1. Find the characteristic polynomial.
You can do this step using the following command
[V, lam] = eig(A)
which produces for the variable lam the eigenvalues with repeats, allowing one to easily deduce the following for the characteristic polynomial.
$chi_A(lambda) = (lambda - 3)^4(lambda + 1)$.
STEP 2. Find the geometric multiplicity for $lambda = 3$.
To do this we must find the nullity, which is equal to $5 - textrank(A - 3I)$ by the rank-nullity theorem. Commanding 5 - rank(A - 3 * eye(5)) gives $2$.
STEP 3. Find the geometric multiplicity for $lambda = -1$.
To do this we must find the nullity, which is equal to $5 - textrank(A + I)$ by the rank-nullity theorem. Commanding 5 - rank(A + eye(5)) gives $1$.
STEP 4. Make a table and try to guess the Jordan form.
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
3 & 4 & 2 \
-1 & 1 & 1
endarray
$$
However in this case we cannot guess, because are two Jordan blocks and their dimensions sum to 4. There are two ways to split 4 into two integers and we don't know which one is it.
How do I proceed? If you could, I would greatly appreciate it if you use the step structure I had been using throughout this post. Ideally with Octave commands.
linear-algebra jordan-normal-form octave
edited Mar 31 at 0:07
idriskameni
751321
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
$endgroup$
$begingroup$
do me a favor, what is the minimal polynomial? If I write it as $(lambda - 3)^k (lambda +1),$ what is the correct $k ; ? ;$
$endgroup$
– Will Jagy
Mar 30 at 23:55
$begingroup$
@WillJagy is it $(lambda - 3)^2(lambda + 1)$?
$endgroup$
– ErotemeObelus
Mar 30 at 23:56
$begingroup$
I don't now, it should be the lowest degree that still gives the zero matrix when your matrix $A$ is the argument..ummm, is $(A - 3I)(A+I)= 0 ??$ If not, what about $(A - 3I)^2(A+I)= 0 ??$
$endgroup$
– Will Jagy
Mar 30 at 23:59
$begingroup$
@WillJagy I think that in order to find the minimal polynomial, you have to find the change-of-basis matrix. And to find the change-of-basis matrix, you have to do something regarding generalized eigenvectors that I don't remember. I think there are three more steps required. Once you have JCF, it is easy to inspect the matrix and determine the minimal polynomial.
$endgroup$
– ErotemeObelus
Mar 31 at 0:02
$begingroup$
Actually, no. Cayley-Hamilton says that $(A-3I)^4 (A+I) = 0.$ The minimal polynomial is with the lowest exponent $k$ such that $(A-3I)^k (A+I) = 0.$ One theorem is that $k geq 1,$ so the possibilities to check are $1 leq k leq 4.$ Once you find $k,$ there is a clear path to finding the Jordan form.
$endgroup$
– Will Jagy
Mar 31 at 0:08
|
show 2 more comments
$begingroup$
I know how to begin the procedure but I don't know how to finish it. Let's start with an example (sorry for it being so unwieldy).
Let
$$A =beginpmatrix
177& 548& 271& -548& -356\ 19& 63& 14& -79& -23\ 8& 24& 17& -20& -20\ 42& 132& 55& -141& -76\ 56& 176& 80& -184& -105endpmatrix$$
Find the Jordan canonical form of A and the change of basis matrix.
STEP 1. Find the characteristic polynomial.
You can do this step using the following command
[V, lam] = eig(A)
which produces for the variable lam the eigenvalues with repeats, allowing one to easily deduce the following for the characteristic polynomial.
$chi_A(lambda) = (lambda - 3)^4(lambda + 1)$.
STEP 2. Find the geometric multiplicity for $lambda = 3$.
To do this we must find the nullity, which is equal to $5 - textrank(A - 3I)$ by the rank-nullity theorem. Commanding 5 - rank(A - 3 * eye(5)) gives $2$.
STEP 3. Find the geometric multiplicity for $lambda = -1$.
To do this we must find the nullity, which is equal to $5 - textrank(A + I)$ by the rank-nullity theorem. Commanding 5 - rank(A + eye(5)) gives $1$.
STEP 4. Make a table and try to guess the Jordan form.
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
3 & 4 & 2 \
-1 & 1 & 1
endarray
$$
However in this case we cannot guess, because are two Jordan blocks and their dimensions sum to 4. There are two ways to split 4 into two integers and we don't know which one is it.
How do I proceed? If you could, I would greatly appreciate it if you use the step structure I had been using throughout this post. Ideally with Octave commands.
linear-algebra jordan-normal-form octave
edited Mar 31 at 0:07
idriskameni
751321
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
$endgroup$
I know how to begin the procedure but I don't know how to finish it. Let's start with an example (sorry for it being so unwieldy).
Let
$$A =beginpmatrix
177& 548& 271& -548& -356\ 19& 63& 14& -79& -23\ 8& 24& 17& -20& -20\ 42& 132& 55& -141& -76\ 56& 176& 80& -184& -105endpmatrix$$
Find the Jordan canonical form of A and the change of basis matrix.
STEP 1. Find the characteristic polynomial.
You can do this step using the following command
[V, lam] = eig(A)
which produces for the variable lam the eigenvalues with repeats, allowing one to easily deduce the following for the characteristic polynomial.
$chi_A(lambda) = (lambda - 3)^4(lambda + 1)$.
STEP 2. Find the geometric multiplicity for $lambda = 3$.
To do this we must find the nullity, which is equal to $5 - textrank(A - 3I)$ by the rank-nullity theorem. Commanding 5 - rank(A - 3 * eye(5)) gives $2$.
STEP 3. Find the geometric multiplicity for $lambda = -1$.
To do this we must find the nullity, which is equal to $5 - textrank(A + I)$ by the rank-nullity theorem. Commanding 5 - rank(A + eye(5)) gives $1$.
STEP 4. Make a table and try to guess the Jordan form.
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
3 & 4 & 2 \
-1 & 1 & 1
endarray
$$
However in this case we cannot guess, because are two Jordan blocks and their dimensions sum to 4. There are two ways to split 4 into two integers and we don't know which one is it.
How do I proceed? If you could, I would greatly appreciate it if you use the step structure I had been using throughout this post. Ideally with Octave commands.
linear-algebra jordan-normal-form octave
linear-algebra jordan-normal-form octave
edited Mar 31 at 0:07
idriskameni
751321
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
edited Mar 31 at 0:07
idriskameni
751321
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
edited Mar 31 at 0:07
idriskameni
751321
edited Mar 31 at 0:07
idriskameni
751321
edited Mar 31 at 0:07
idriskameni
751321
751321
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
asked Mar 30 at 23:52
ErotemeObelusErotemeObelus
681722
681722
$begingroup$
do me a favor, what is the minimal polynomial? If I write it as $(lambda - 3)^k (lambda +1),$ what is the correct $k ; ? ;$
$endgroup$
– Will Jagy
Mar 30 at 23:55
$begingroup$
@WillJagy is it $(lambda - 3)^2(lambda + 1)$?
$endgroup$
– ErotemeObelus
Mar 30 at 23:56
$begingroup$
I don't now, it should be the lowest degree that still gives the zero matrix when your matrix $A$ is the argument..ummm, is $(A - 3I)(A+I)= 0 ??$ If not, what about $(A - 3I)^2(A+I)= 0 ??$
$endgroup$
– Will Jagy
Mar 30 at 23:59
$begingroup$
@WillJagy I think that in order to find the minimal polynomial, you have to find the change-of-basis matrix. And to find the change-of-basis matrix, you have to do something regarding generalized eigenvectors that I don't remember. I think there are three more steps required. Once you have JCF, it is easy to inspect the matrix and determine the minimal polynomial.
$endgroup$
– ErotemeObelus
Mar 31 at 0:02
$begingroup$
Actually, no. Cayley-Hamilton says that $(A-3I)^4 (A+I) = 0.$ The minimal polynomial is with the lowest exponent $k$ such that $(A-3I)^k (A+I) = 0.$ One theorem is that $k geq 1,$ so the possibilities to check are $1 leq k leq 4.$ Once you find $k,$ there is a clear path to finding the Jordan form.
$endgroup$
– Will Jagy
Mar 31 at 0:08
|
show 2 more comments
$begingroup$
do me a favor, what is the minimal polynomial? If I write it as $(lambda - 3)^k (lambda +1),$ what is the correct $k ; ? ;$
$endgroup$
– Will Jagy
Mar 30 at 23:55
$begingroup$
@WillJagy is it $(lambda - 3)^2(lambda + 1)$?
$endgroup$
– ErotemeObelus
Mar 30 at 23:56
$begingroup$
I don't now, it should be the lowest degree that still gives the zero matrix when your matrix $A$ is the argument..ummm, is $(A - 3I)(A+I)= 0 ??$ If not, what about $(A - 3I)^2(A+I)= 0 ??$
$endgroup$
– Will Jagy
Mar 30 at 23:59
$begingroup$
@WillJagy I think that in order to find the minimal polynomial, you have to find the change-of-basis matrix. And to find the change-of-basis matrix, you have to do something regarding generalized eigenvectors that I don't remember. I think there are three more steps required. Once you have JCF, it is easy to inspect the matrix and determine the minimal polynomial.
$endgroup$
– ErotemeObelus
Mar 31 at 0:02
$begingroup$
Actually, no. Cayley-Hamilton says that $(A-3I)^4 (A+I) = 0.$ The minimal polynomial is with the lowest exponent $k$ such that $(A-3I)^k (A+I) = 0.$ One theorem is that $k geq 1,$ so the possibilities to check are $1 leq k leq 4.$ Once you find $k,$ there is a clear path to finding the Jordan form.
$endgroup$
– Will Jagy
Mar 31 at 0:08
$begingroup$
do me a favor, what is the minimal polynomial? If I write it as $(lambda - 3)^k (lambda +1),$ what is the correct $k ; ? ;$
$endgroup$
– Will Jagy
Mar 30 at 23:55
$begingroup$
do me a favor, what is the minimal polynomial? If I write it as $(lambda - 3)^k (lambda +1),$ what is the correct $k ; ? ;$
$endgroup$
– Will Jagy
Mar 30 at 23:55
$begingroup$
@WillJagy is it $(lambda - 3)^2(lambda + 1)$?
$endgroup$
– ErotemeObelus
Mar 30 at 23:56
$begingroup$
@WillJagy is it $(lambda - 3)^2(lambda + 1)$?
$endgroup$
– ErotemeObelus
Mar 30 at 23:56
$begingroup$
I don't now, it should be the lowest degree that still gives the zero matrix when your matrix $A$ is the argument..ummm, is $(A - 3I)(A+I)= 0 ??$ If not, what about $(A - 3I)^2(A+I)= 0 ??$
$endgroup$
– Will Jagy
Mar 30 at 23:59
$begingroup$
I don't now, it should be the lowest degree that still gives the zero matrix when your matrix $A$ is the argument..ummm, is $(A - 3I)(A+I)= 0 ??$ If not, what about $(A - 3I)^2(A+I)= 0 ??$
$endgroup$
– Will Jagy
Mar 30 at 23:59
$begingroup$
@WillJagy I think that in order to find the minimal polynomial, you have to find the change-of-basis matrix. And to find the change-of-basis matrix, you have to do something regarding generalized eigenvectors that I don't remember. I think there are three more steps required. Once you have JCF, it is easy to inspect the matrix and determine the minimal polynomial.
$endgroup$
– ErotemeObelus
Mar 31 at 0:02
$begingroup$
@WillJagy I think that in order to find the minimal polynomial, you have to find the change-of-basis matrix. And to find the change-of-basis matrix, you have to do something regarding generalized eigenvectors that I don't remember. I think there are three more steps required. Once you have JCF, it is easy to inspect the matrix and determine the minimal polynomial.
$endgroup$
– ErotemeObelus
Mar 31 at 0:02
$begingroup$
Actually, no. Cayley-Hamilton says that $(A-3I)^4 (A+I) = 0.$ The minimal polynomial is with the lowest exponent $k$ such that $(A-3I)^k (A+I) = 0.$ One theorem is that $k geq 1,$ so the possibilities to check are $1 leq k leq 4.$ Once you find $k,$ there is a clear path to finding the Jordan form.
$endgroup$
– Will Jagy
Mar 31 at 0:08
$begingroup$
Actually, no. Cayley-Hamilton says that $(A-3I)^4 (A+I) = 0.$ The minimal polynomial is with the lowest exponent $k$ such that $(A-3I)^k (A+I) = 0.$ One theorem is that $k geq 1,$ so the possibilities to check are $1 leq k leq 4.$ Once you find $k,$ there is a clear path to finding the Jordan form.
$endgroup$
– Will Jagy
Mar 31 at 0:08
|
show 2 more comments
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$begingroup$
do me a favor, what is the minimal polynomial? If I write it as $(lambda - 3)^k (lambda +1),$ what is the correct $k ; ? ;$
$endgroup$
– Will Jagy
Mar 30 at 23:55
$begingroup$
@WillJagy is it $(lambda - 3)^2(lambda + 1)$?
$endgroup$
– ErotemeObelus
Mar 30 at 23:56
$begingroup$
I don't now, it should be the lowest degree that still gives the zero matrix when your matrix $A$ is the argument..ummm, is $(A - 3I)(A+I)= 0 ??$ If not, what about $(A - 3I)^2(A+I)= 0 ??$
$endgroup$
– Will Jagy
Mar 30 at 23:59
$begingroup$
@WillJagy I think that in order to find the minimal polynomial, you have to find the change-of-basis matrix. And to find the change-of-basis matrix, you have to do something regarding generalized eigenvectors that I don't remember. I think there are three more steps required. Once you have JCF, it is easy to inspect the matrix and determine the minimal polynomial.
$endgroup$
– ErotemeObelus
Mar 31 at 0:02
$begingroup$
Actually, no. Cayley-Hamilton says that $(A-3I)^4 (A+I) = 0.$ The minimal polynomial is with the lowest exponent $k$ such that $(A-3I)^k (A+I) = 0.$ One theorem is that $k geq 1,$ so the possibilities to check are $1 leq k leq 4.$ Once you find $k,$ there is a clear path to finding the Jordan form.
$endgroup$
– Will Jagy
Mar 31 at 0:08