cyclic vectors- cyclic subspaces
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Let $T$ be a linear operator on ${mathbb R}^3$, which is represented by standard ordered basis, as follows:
$$T(e_1) = 2 e_1,$$
$$T(e_2) = 2 e_2,$$
$$T( e_3) = -e_3.$$
I have to prove that $T$ has no cyclic vector. Also, to find what is the $T$-cyclic subspace generated by the $(1, -1, 3 )$ ?
Please suggest !
linear-algebra
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
asked Feb 27 '12 at 11:00
preeti
587720
add a comment |
up vote
2
down vote
favorite
Let $T$ be a linear operator on ${mathbb R}^3$, which is represented by standard ordered basis, as follows:
$$T(e_1) = 2 e_1,$$
$$T(e_2) = 2 e_2,$$
$$T( e_3) = -e_3.$$
I have to prove that $T$ has no cyclic vector. Also, to find what is the $T$-cyclic subspace generated by the $(1, -1, 3 )$ ?
Please suggest !
linear-algebra
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
asked Feb 27 '12 at 11:00
preeti
587720
If a matrix is brought into diagonal (or triangular) form, it's cyclic vectors are easy to detect: it is necessary and sufficient that for each distinct eigenvalue the projection of the vector on the generalised eigenspace be a cyclic vector for the subspace, where projecting in this case means extracting the appropriate coordinates. Your example is quite easy in that the matrix is already diagonal. Only multiple eigenvalues need attention.
– Marc van Leeuwen
Feb 27 '12 at 11:23
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $T$ be a linear operator on ${mathbb R}^3$, which is represented by standard ordered basis, as follows:
$$T(e_1) = 2 e_1,$$
$$T(e_2) = 2 e_2,$$
$$T( e_3) = -e_3.$$
I have to prove that $T$ has no cyclic vector. Also, to find what is the $T$-cyclic subspace generated by the $(1, -1, 3 )$ ?
Please suggest !
linear-algebra
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
asked Feb 27 '12 at 11:00
preeti
587720
Let $T$ be a linear operator on ${mathbb R}^3$, which is represented by standard ordered basis, as follows:
$$T(e_1) = 2 e_1,$$
$$T(e_2) = 2 e_2,$$
$$T( e_3) = -e_3.$$
I have to prove that $T$ has no cyclic vector. Also, to find what is the $T$-cyclic subspace generated by the $(1, -1, 3 )$ ?
Please suggest !
linear-algebra
linear-algebra
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
asked Feb 27 '12 at 11:00
preeti
587720
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
asked Feb 27 '12 at 11:00
preeti
587720
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
edited Feb 27 '12 at 11:11
Christian Blatter
171k7111325
171k7111325
asked Feb 27 '12 at 11:00
preeti
587720
asked Feb 27 '12 at 11:00
preeti
587720
asked Feb 27 '12 at 11:00
preeti
587720
587720
If a matrix is brought into diagonal (or triangular) form, it's cyclic vectors are easy to detect: it is necessary and sufficient that for each distinct eigenvalue the projection of the vector on the generalised eigenspace be a cyclic vector for the subspace, where projecting in this case means extracting the appropriate coordinates. Your example is quite easy in that the matrix is already diagonal. Only multiple eigenvalues need attention.
– Marc van Leeuwen
Feb 27 '12 at 11:23
add a comment |
If a matrix is brought into diagonal (or triangular) form, it's cyclic vectors are easy to detect: it is necessary and sufficient that for each distinct eigenvalue the projection of the vector on the generalised eigenspace be a cyclic vector for the subspace, where projecting in this case means extracting the appropriate coordinates. Your example is quite easy in that the matrix is already diagonal. Only multiple eigenvalues need attention.
– Marc van Leeuwen
Feb 27 '12 at 11:23
If a matrix is brought into diagonal (or triangular) form, it's cyclic vectors are easy to detect: it is necessary and sufficient that for each distinct eigenvalue the projection of the vector on the generalised eigenspace be a cyclic vector for the subspace, where projecting in this case means extracting the appropriate coordinates. Your example is quite easy in that the matrix is already diagonal. Only multiple eigenvalues need attention.
– Marc van Leeuwen
Feb 27 '12 at 11:23
If a matrix is brought into diagonal (or triangular) form, it's cyclic vectors are easy to detect: it is necessary and sufficient that for each distinct eigenvalue the projection of the vector on the generalised eigenspace be a cyclic vector for the subspace, where projecting in this case means extracting the appropriate coordinates. Your example is quite easy in that the matrix is already diagonal. Only multiple eigenvalues need attention.
– Marc van Leeuwen
Feb 27 '12 at 11:23
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$. Both are not same. Hence $T$ doesn't have cyclic vector by following theorem:
Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....
Proof:Suppose there exists a cyclic vector $v$ for $T$, that is we have $vin V$ such that ${v, Tv,..T^{n-1}v}$ span $V$. Then matrix representation of $T$ will be some companion matrix , whose minimal and characteristic polynomial are same.
Now conversely, if minimal and charteristic polynomial are same, then we have a minimal polynomial which is of degree $n$. Take $vneq 0 $, let minimal polynomial $p(x)= a_0+a_1x+...+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider ${v, Tv, T^2v,.. T^{n-1}v}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear algebra by Hoffman-Kunze.
Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence T-cyclic subspace generated by the $(1,−1,3)$ = Linear span of ${(1,-1,3), (2,-2,-3)}$
edited Nov 23 at 15:33
Thomas Shelby
932116
answered Feb 27 '12 at 11:13
zapkm
1,461915
Thanks a lot for the help.
– preeti
Feb 27 '12 at 11:52
@zapkm $+1$, could u tell me which book did u get this theorem?
– Wow
May 10 '13 at 18:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$. Both are not same. Hence $T$ doesn't have cyclic vector by following theorem:
Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....
Proof:Suppose there exists a cyclic vector $v$ for $T$, that is we have $vin V$ such that ${v, Tv,..T^{n-1}v}$ span $V$. Then matrix representation of $T$ will be some companion matrix , whose minimal and characteristic polynomial are same.
Now conversely, if minimal and charteristic polynomial are same, then we have a minimal polynomial which is of degree $n$. Take $vneq 0 $, let minimal polynomial $p(x)= a_0+a_1x+...+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider ${v, Tv, T^2v,.. T^{n-1}v}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear algebra by Hoffman-Kunze.
Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence T-cyclic subspace generated by the $(1,−1,3)$ = Linear span of ${(1,-1,3), (2,-2,-3)}$
edited Nov 23 at 15:33
Thomas Shelby
932116
answered Feb 27 '12 at 11:13
zapkm
1,461915
Thanks a lot for the help.
– preeti
Feb 27 '12 at 11:52
@zapkm $+1$, could u tell me which book did u get this theorem?
– Wow
May 10 '13 at 18:07
add a comment |
up vote
5
down vote
accepted
Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$. Both are not same. Hence $T$ doesn't have cyclic vector by following theorem:
Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....
Proof:Suppose there exists a cyclic vector $v$ for $T$, that is we have $vin V$ such that ${v, Tv,..T^{n-1}v}$ span $V$. Then matrix representation of $T$ will be some companion matrix , whose minimal and characteristic polynomial are same.
Now conversely, if minimal and charteristic polynomial are same, then we have a minimal polynomial which is of degree $n$. Take $vneq 0 $, let minimal polynomial $p(x)= a_0+a_1x+...+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider ${v, Tv, T^2v,.. T^{n-1}v}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear algebra by Hoffman-Kunze.
Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence T-cyclic subspace generated by the $(1,−1,3)$ = Linear span of ${(1,-1,3), (2,-2,-3)}$
edited Nov 23 at 15:33
Thomas Shelby
932116
answered Feb 27 '12 at 11:13
zapkm
1,461915
Thanks a lot for the help.
– preeti
Feb 27 '12 at 11:52
@zapkm $+1$, could u tell me which book did u get this theorem?
– Wow
May 10 '13 at 18:07
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$. Both are not same. Hence $T$ doesn't have cyclic vector by following theorem:
Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....
Proof:Suppose there exists a cyclic vector $v$ for $T$, that is we have $vin V$ such that ${v, Tv,..T^{n-1}v}$ span $V$. Then matrix representation of $T$ will be some companion matrix , whose minimal and characteristic polynomial are same.
Now conversely, if minimal and charteristic polynomial are same, then we have a minimal polynomial which is of degree $n$. Take $vneq 0 $, let minimal polynomial $p(x)= a_0+a_1x+...+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider ${v, Tv, T^2v,.. T^{n-1}v}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear algebra by Hoffman-Kunze.
Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence T-cyclic subspace generated by the $(1,−1,3)$ = Linear span of ${(1,-1,3), (2,-2,-3)}$
edited Nov 23 at 15:33
Thomas Shelby
932116
answered Feb 27 '12 at 11:13
zapkm
1,461915
Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$. Both are not same. Hence $T$ doesn't have cyclic vector by following theorem:
Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....
Proof:Suppose there exists a cyclic vector $v$ for $T$, that is we have $vin V$ such that ${v, Tv,..T^{n-1}v}$ span $V$. Then matrix representation of $T$ will be some companion matrix , whose minimal and characteristic polynomial are same.
Now conversely, if minimal and charteristic polynomial are same, then we have a minimal polynomial which is of degree $n$. Take $vneq 0 $, let minimal polynomial $p(x)= a_0+a_1x+...+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider ${v, Tv, T^2v,.. T^{n-1}v}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear algebra by Hoffman-Kunze.
Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence T-cyclic subspace generated by the $(1,−1,3)$ = Linear span of ${(1,-1,3), (2,-2,-3)}$
edited Nov 23 at 15:33
Thomas Shelby
932116
answered Feb 27 '12 at 11:13
zapkm
1,461915
edited Nov 23 at 15:33
Thomas Shelby
932116
edited Nov 23 at 15:33
Thomas Shelby
932116
edited Nov 23 at 15:33
Thomas Shelby
932116
932116
answered Feb 27 '12 at 11:13
zapkm
1,461915
answered Feb 27 '12 at 11:13
zapkm
1,461915
answered Feb 27 '12 at 11:13
zapkm
1,461915
1,461915
Thanks a lot for the help.
– preeti
Feb 27 '12 at 11:52
@zapkm $+1$, could u tell me which book did u get this theorem?
– Wow
May 10 '13 at 18:07
add a comment |
Thanks a lot for the help.
– preeti
Feb 27 '12 at 11:52
@zapkm $+1$, could u tell me which book did u get this theorem?
– Wow
May 10 '13 at 18:07
Thanks a lot for the help.
– preeti
Feb 27 '12 at 11:52
Thanks a lot for the help.
– preeti
Feb 27 '12 at 11:52
@zapkm $+1$, could u tell me which book did u get this theorem?
– Wow
May 10 '13 at 18:07
@zapkm $+1$, could u tell me which book did u get this theorem?
– Wow
May 10 '13 at 18:07
add a comment |
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If a matrix is brought into diagonal (or triangular) form, it's cyclic vectors are easy to detect: it is necessary and sufficient that for each distinct eigenvalue the projection of the vector on the generalised eigenspace be a cyclic vector for the subspace, where projecting in this case means extracting the appropriate coordinates. Your example is quite easy in that the matrix is already diagonal. Only multiple eigenvalues need attention.
– Marc van Leeuwen
Feb 27 '12 at 11:23