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Find the dimension of the Eigen-space corresponding to the Eigen value $ lambda $ with respect to $ vec v $
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Let $ V $ be a finite dimensional vector space over the field $ mathbbC $ and $ T: V to V $ be a linear transformation.
Suppose $ lambda $ is a Eigenvalue of $ T $.
Let the characteristics polynomial and minimal polynomial of $ T $ be respectively given by $ chi(lambda)=(z-lambda)^m $ and $ m(lambda)=z-lambda $ .
If $ vec v $ be the Jordan basis of $ T $ , then find $ dim E(lambda , vec v) $ or the dimension of the Eigen-space corresponding to the Eigen value $ lambda $ with respect to $ vec v $.
Also find the Jordan matrix $ mathcalM (T) $
Answer:
Since the characteristic polynomial of $ T $ is $ chi(lambda)=(z-lambda)^m $ , we have $ dim (V)=m $ ,
Now the algebraic multiplicity of $ lambda $ is $ m $.
Since the minimal polynomial of $ T $ is $ m(lambda)=z-lambda $ , the geometric multiplicity of $ lambda $ is also $ m $.
Thus the dimension of the Eigen space $ E(lambda) $ of $ lambda $ is given by
$ dim E(lambda)=m $
Am I right so far?
So there is a jordan block of size $ m $ as follows:
$$ mathcalM(T)=beginpmatrix lambda & 0 & 0 & cdots & 0 \ 0 & lambda & 0 & cdots & 0 \ vdots & vdots & & & vdots \ 0 & 0 & & cdots & lambda endpmatrix $$
Am I right so far?
linear-algebra jordan-normal-form
asked Jun 29 at 0:45
yourmath
1,8021617
add a comment |Â
up vote
4
down vote
favorite
Let $ V $ be a finite dimensional vector space over the field $ mathbbC $ and $ T: V to V $ be a linear transformation.
Suppose $ lambda $ is a Eigenvalue of $ T $.
Let the characteristics polynomial and minimal polynomial of $ T $ be respectively given by $ chi(lambda)=(z-lambda)^m $ and $ m(lambda)=z-lambda $ .
If $ vec v $ be the Jordan basis of $ T $ , then find $ dim E(lambda , vec v) $ or the dimension of the Eigen-space corresponding to the Eigen value $ lambda $ with respect to $ vec v $.
Also find the Jordan matrix $ mathcalM (T) $
Answer:
Since the characteristic polynomial of $ T $ is $ chi(lambda)=(z-lambda)^m $ , we have $ dim (V)=m $ ,
Now the algebraic multiplicity of $ lambda $ is $ m $.
Since the minimal polynomial of $ T $ is $ m(lambda)=z-lambda $ , the geometric multiplicity of $ lambda $ is also $ m $.
Thus the dimension of the Eigen space $ E(lambda) $ of $ lambda $ is given by
$ dim E(lambda)=m $
Am I right so far?
So there is a jordan block of size $ m $ as follows:
$$ mathcalM(T)=beginpmatrix lambda & 0 & 0 & cdots & 0 \ 0 & lambda & 0 & cdots & 0 \ vdots & vdots & & & vdots \ 0 & 0 & & cdots & lambda endpmatrix $$
Am I right so far?
linear-algebra jordan-normal-form
asked Jun 29 at 0:45
yourmath
1,8021617
1
If the minimal polynomial is $z-lambda$, by definition $T-lambda I=0$. Therefore $T=lambda I$. Which means that all vectors are eigenvectors corresponding to the eigevalue $lambda$. All the remaining information is not needed: The characteristic polynomial, the finite-dimensionality of $V$, even that $T$ is linear is not needed. Everything follows from knowing that the minimal polynomial is that simple.
– user569098
Jun 29 at 1:56
@LB_O "that $T$ is linear is not needed." How do you define the minimal polynomial of a non-linear transformation?
– user539887
Jun 29 at 6:57
Can someone answer the question ?
– yourmath
Jun 29 at 7:57
@user539887 Just lookup the definition of minimal polynomial and change nothing. All you need is a function $f:Vto V$ where $V$ is a vector space, or even less, a module over the ring of coefficients of the polynomial.
– user569098
Jun 29 at 10:51
2
@yourmath I already answered the question: Since the minimal polynomial is $z-lambda$, then $T$ is equal to multiplication by $lambda$. Therefore, all vectors are eigenvectors, the dimension of the eigenspace for $lambda$ is the dimension of $V$, and in any basis the matrix of $T$ is diagonal with $lambda$ in the diagonal. In particular, that is its Jordan form.
– user569098
Jun 29 at 10:54
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $ V $ be a finite dimensional vector space over the field $ mathbbC $ and $ T: V to V $ be a linear transformation.
Suppose $ lambda $ is a Eigenvalue of $ T $.
Let the characteristics polynomial and minimal polynomial of $ T $ be respectively given by $ chi(lambda)=(z-lambda)^m $ and $ m(lambda)=z-lambda $ .
If $ vec v $ be the Jordan basis of $ T $ , then find $ dim E(lambda , vec v) $ or the dimension of the Eigen-space corresponding to the Eigen value $ lambda $ with respect to $ vec v $.
Also find the Jordan matrix $ mathcalM (T) $
Answer:
Since the characteristic polynomial of $ T $ is $ chi(lambda)=(z-lambda)^m $ , we have $ dim (V)=m $ ,
Now the algebraic multiplicity of $ lambda $ is $ m $.
Since the minimal polynomial of $ T $ is $ m(lambda)=z-lambda $ , the geometric multiplicity of $ lambda $ is also $ m $.
Thus the dimension of the Eigen space $ E(lambda) $ of $ lambda $ is given by
$ dim E(lambda)=m $
Am I right so far?
So there is a jordan block of size $ m $ as follows:
$$ mathcalM(T)=beginpmatrix lambda & 0 & 0 & cdots & 0 \ 0 & lambda & 0 & cdots & 0 \ vdots & vdots & & & vdots \ 0 & 0 & & cdots & lambda endpmatrix $$
Am I right so far?
linear-algebra jordan-normal-form
asked Jun 29 at 0:45
yourmath
1,8021617
Let $ V $ be a finite dimensional vector space over the field $ mathbbC $ and $ T: V to V $ be a linear transformation.
Suppose $ lambda $ is a Eigenvalue of $ T $.
Let the characteristics polynomial and minimal polynomial of $ T $ be respectively given by $ chi(lambda)=(z-lambda)^m $ and $ m(lambda)=z-lambda $ .
If $ vec v $ be the Jordan basis of $ T $ , then find $ dim E(lambda , vec v) $ or the dimension of the Eigen-space corresponding to the Eigen value $ lambda $ with respect to $ vec v $.
Also find the Jordan matrix $ mathcalM (T) $
Answer:
Since the characteristic polynomial of $ T $ is $ chi(lambda)=(z-lambda)^m $ , we have $ dim (V)=m $ ,
Now the algebraic multiplicity of $ lambda $ is $ m $.
Since the minimal polynomial of $ T $ is $ m(lambda)=z-lambda $ , the geometric multiplicity of $ lambda $ is also $ m $.
Thus the dimension of the Eigen space $ E(lambda) $ of $ lambda $ is given by
$ dim E(lambda)=m $
Am I right so far?
So there is a jordan block of size $ m $ as follows:
$$ mathcalM(T)=beginpmatrix lambda & 0 & 0 & cdots & 0 \ 0 & lambda & 0 & cdots & 0 \ vdots & vdots & & & vdots \ 0 & 0 & & cdots & lambda endpmatrix $$
Am I right so far?
linear-algebra jordan-normal-form
asked Jun 29 at 0:45
yourmath
1,8021617
asked Jun 29 at 0:45
yourmath
1,8021617
asked Jun 29 at 0:45
yourmath
1,8021617
asked Jun 29 at 0:45
yourmath
1,8021617
1,8021617
1
If the minimal polynomial is $z-lambda$, by definition $T-lambda I=0$. Therefore $T=lambda I$. Which means that all vectors are eigenvectors corresponding to the eigevalue $lambda$. All the remaining information is not needed: The characteristic polynomial, the finite-dimensionality of $V$, even that $T$ is linear is not needed. Everything follows from knowing that the minimal polynomial is that simple.
– user569098
Jun 29 at 1:56
@LB_O "that $T$ is linear is not needed." How do you define the minimal polynomial of a non-linear transformation?
– user539887
Jun 29 at 6:57
Can someone answer the question ?
– yourmath
Jun 29 at 7:57
@user539887 Just lookup the definition of minimal polynomial and change nothing. All you need is a function $f:Vto V$ where $V$ is a vector space, or even less, a module over the ring of coefficients of the polynomial.
– user569098
Jun 29 at 10:51
2
@yourmath I already answered the question: Since the minimal polynomial is $z-lambda$, then $T$ is equal to multiplication by $lambda$. Therefore, all vectors are eigenvectors, the dimension of the eigenspace for $lambda$ is the dimension of $V$, and in any basis the matrix of $T$ is diagonal with $lambda$ in the diagonal. In particular, that is its Jordan form.
– user569098
Jun 29 at 10:54
add a comment |Â
1
If the minimal polynomial is $z-lambda$, by definition $T-lambda I=0$. Therefore $T=lambda I$. Which means that all vectors are eigenvectors corresponding to the eigevalue $lambda$. All the remaining information is not needed: The characteristic polynomial, the finite-dimensionality of $V$, even that $T$ is linear is not needed. Everything follows from knowing that the minimal polynomial is that simple.
– user569098
Jun 29 at 1:56
@LB_O "that $T$ is linear is not needed." How do you define the minimal polynomial of a non-linear transformation?
– user539887
Jun 29 at 6:57
Can someone answer the question ?
– yourmath
Jun 29 at 7:57
@user539887 Just lookup the definition of minimal polynomial and change nothing. All you need is a function $f:Vto V$ where $V$ is a vector space, or even less, a module over the ring of coefficients of the polynomial.
– user569098
Jun 29 at 10:51
2
@yourmath I already answered the question: Since the minimal polynomial is $z-lambda$, then $T$ is equal to multiplication by $lambda$. Therefore, all vectors are eigenvectors, the dimension of the eigenspace for $lambda$ is the dimension of $V$, and in any basis the matrix of $T$ is diagonal with $lambda$ in the diagonal. In particular, that is its Jordan form.
– user569098
Jun 29 at 10:54
1
1
If the minimal polynomial is $z-lambda$, by definition $T-lambda I=0$. Therefore $T=lambda I$. Which means that all vectors are eigenvectors corresponding to the eigevalue $lambda$. All the remaining information is not needed: The characteristic polynomial, the finite-dimensionality of $V$, even that $T$ is linear is not needed. Everything follows from knowing that the minimal polynomial is that simple.
– user569098
Jun 29 at 1:56
If the minimal polynomial is $z-lambda$, by definition $T-lambda I=0$. Therefore $T=lambda I$. Which means that all vectors are eigenvectors corresponding to the eigevalue $lambda$. All the remaining information is not needed: The characteristic polynomial, the finite-dimensionality of $V$, even that $T$ is linear is not needed. Everything follows from knowing that the minimal polynomial is that simple.
– user569098
Jun 29 at 1:56
@LB_O "that $T$ is linear is not needed." How do you define the minimal polynomial of a non-linear transformation?
– user539887
Jun 29 at 6:57
@LB_O "that $T$ is linear is not needed." How do you define the minimal polynomial of a non-linear transformation?
– user539887
Jun 29 at 6:57
Can someone answer the question ?
– yourmath
Jun 29 at 7:57
Can someone answer the question ?
– yourmath
Jun 29 at 7:57
@user539887 Just lookup the definition of minimal polynomial and change nothing. All you need is a function $f:Vto V$ where $V$ is a vector space, or even less, a module over the ring of coefficients of the polynomial.
– user569098
Jun 29 at 10:51
@user539887 Just lookup the definition of minimal polynomial and change nothing. All you need is a function $f:Vto V$ where $V$ is a vector space, or even less, a module over the ring of coefficients of the polynomial.
– user569098
Jun 29 at 10:51
2
2
@yourmath I already answered the question: Since the minimal polynomial is $z-lambda$, then $T$ is equal to multiplication by $lambda$. Therefore, all vectors are eigenvectors, the dimension of the eigenspace for $lambda$ is the dimension of $V$, and in any basis the matrix of $T$ is diagonal with $lambda$ in the diagonal. In particular, that is its Jordan form.
– user569098
Jun 29 at 10:54
@yourmath I already answered the question: Since the minimal polynomial is $z-lambda$, then $T$ is equal to multiplication by $lambda$. Therefore, all vectors are eigenvectors, the dimension of the eigenspace for $lambda$ is the dimension of $V$, and in any basis the matrix of $T$ is diagonal with $lambda$ in the diagonal. In particular, that is its Jordan form.
– user569098
Jun 29 at 10:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Yes, you are right. For justification, see Wikipedia subtopic "Complex matrices" in the "Jordan normal form" article.
answered Aug 6 at 16:55
Maurice P
1,1501630
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
Yes, you are right. For justification, see Wikipedia subtopic "Complex matrices" in the "Jordan normal form" article.
answered Aug 6 at 16:55
Maurice P
1,1501630
add a comment |Â
up vote
1
down vote
Yes, you are right. For justification, see Wikipedia subtopic "Complex matrices" in the "Jordan normal form" article.
answered Aug 6 at 16:55
Maurice P
1,1501630
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, you are right. For justification, see Wikipedia subtopic "Complex matrices" in the "Jordan normal form" article.
answered Aug 6 at 16:55
Maurice P
1,1501630
Yes, you are right. For justification, see Wikipedia subtopic "Complex matrices" in the "Jordan normal form" article.
answered Aug 6 at 16:55
Maurice P
1,1501630
answered Aug 6 at 16:55
Maurice P
1,1501630
answered Aug 6 at 16:55
Maurice P
1,1501630
answered Aug 6 at 16:55
Maurice P
1,1501630
1,1501630
add a comment |Â
add a comment |Â
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1
If the minimal polynomial is $z-lambda$, by definition $T-lambda I=0$. Therefore $T=lambda I$. Which means that all vectors are eigenvectors corresponding to the eigevalue $lambda$. All the remaining information is not needed: The characteristic polynomial, the finite-dimensionality of $V$, even that $T$ is linear is not needed. Everything follows from knowing that the minimal polynomial is that simple.
– user569098
Jun 29 at 1:56
@LB_O "that $T$ is linear is not needed." How do you define the minimal polynomial of a non-linear transformation?
– user539887
Jun 29 at 6:57
Can someone answer the question ?
– yourmath
Jun 29 at 7:57
@user539887 Just lookup the definition of minimal polynomial and change nothing. All you need is a function $f:Vto V$ where $V$ is a vector space, or even less, a module over the ring of coefficients of the polynomial.
– user569098
Jun 29 at 10:51
2
@yourmath I already answered the question: Since the minimal polynomial is $z-lambda$, then $T$ is equal to multiplication by $lambda$. Therefore, all vectors are eigenvectors, the dimension of the eigenspace for $lambda$ is the dimension of $V$, and in any basis the matrix of $T$ is diagonal with $lambda$ in the diagonal. In particular, that is its Jordan form.
– user569098
Jun 29 at 10:54