Does transformation invariance of the range and null space imply commutativity?
up vote
1
down vote
favorite
Suppose $R(U)$ and $N(U)$ (the range and null space of a linear transformation $U$, respectively) are $T$-invariant ($T$ linear) subspaces of some vector space $V$. Does this imply $UT=TU$?
I've proven the converse, that $UT=TU$ implies the $T$-invariance of the range and null space. However, I'm a bit stumped trying to prove or come up with a counterexample for the above.
For instance, I can see that when something is in the null space of $U$ that $UT=TU=0$ trivially, but does this extend to the range as well?
linear-algebra vector-spaces linear-transformations invariant-subspace
asked Nov 15 at 1:59
notadoctor
867
|
show 3 more comments
up vote
1
down vote
favorite
Suppose $R(U)$ and $N(U)$ (the range and null space of a linear transformation $U$, respectively) are $T$-invariant ($T$ linear) subspaces of some vector space $V$. Does this imply $UT=TU$?
I've proven the converse, that $UT=TU$ implies the $T$-invariance of the range and null space. However, I'm a bit stumped trying to prove or come up with a counterexample for the above.
For instance, I can see that when something is in the null space of $U$ that $UT=TU=0$ trivially, but does this extend to the range as well?
linear-algebra vector-spaces linear-transformations invariant-subspace
asked Nov 15 at 1:59
notadoctor
867
2
If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$.
– user1551
Nov 15 at 2:14
Is the converse of this true? That is, does the fact that $R(U)$ and $N(U)$ are $T$-invariant imply that $U$ is invertible? That would then prove that $UT=TU$, correct?
– notadoctor
Nov 15 at 2:19
Cool problem. Where did you get this problem?
– mathnoob
Nov 15 at 2:23
2
@justadampaul No. Consider $U=pmatrix{1\ &2\ &&0}, T=pmatrix{0&1\ 1&0\ &&1}$. Then $R(U)={(ast,ast,0)^T}$ and $N(U)={(0,0,ast)^T}$ are $T$-invariant, but $U$ is singular and $UTne TU$.
– user1551
Nov 15 at 2:33
1
It doesn't have to be non-invertible. user1551's first comment suggests the opposite. Consider two rotations in 3D; both are invertible and have trivial nullspace and range, but rotations are not commutative in general.
– mr_e_man
Nov 15 at 2:50
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $R(U)$ and $N(U)$ (the range and null space of a linear transformation $U$, respectively) are $T$-invariant ($T$ linear) subspaces of some vector space $V$. Does this imply $UT=TU$?
I've proven the converse, that $UT=TU$ implies the $T$-invariance of the range and null space. However, I'm a bit stumped trying to prove or come up with a counterexample for the above.
For instance, I can see that when something is in the null space of $U$ that $UT=TU=0$ trivially, but does this extend to the range as well?
linear-algebra vector-spaces linear-transformations invariant-subspace
asked Nov 15 at 1:59
notadoctor
867
Suppose $R(U)$ and $N(U)$ (the range and null space of a linear transformation $U$, respectively) are $T$-invariant ($T$ linear) subspaces of some vector space $V$. Does this imply $UT=TU$?
I've proven the converse, that $UT=TU$ implies the $T$-invariance of the range and null space. However, I'm a bit stumped trying to prove or come up with a counterexample for the above.
For instance, I can see that when something is in the null space of $U$ that $UT=TU=0$ trivially, but does this extend to the range as well?
linear-algebra vector-spaces linear-transformations invariant-subspace
linear-algebra vector-spaces linear-transformations invariant-subspace
asked Nov 15 at 1:59
notadoctor
867
asked Nov 15 at 1:59
notadoctor
867
edited Nov 15 at 2:33
asked Nov 15 at 1:59
notadoctor
867
asked Nov 15 at 1:59
notadoctor
867
asked Nov 15 at 1:59
notadoctor
867
867
2
If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$.
– user1551
Nov 15 at 2:14
Is the converse of this true? That is, does the fact that $R(U)$ and $N(U)$ are $T$-invariant imply that $U$ is invertible? That would then prove that $UT=TU$, correct?
– notadoctor
Nov 15 at 2:19
Cool problem. Where did you get this problem?
– mathnoob
Nov 15 at 2:23
2
@justadampaul No. Consider $U=pmatrix{1\ &2\ &&0}, T=pmatrix{0&1\ 1&0\ &&1}$. Then $R(U)={(ast,ast,0)^T}$ and $N(U)={(0,0,ast)^T}$ are $T$-invariant, but $U$ is singular and $UTne TU$.
– user1551
Nov 15 at 2:33
1
It doesn't have to be non-invertible. user1551's first comment suggests the opposite. Consider two rotations in 3D; both are invertible and have trivial nullspace and range, but rotations are not commutative in general.
– mr_e_man
Nov 15 at 2:50
|
show 3 more comments
2
If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$.
– user1551
Nov 15 at 2:14
Is the converse of this true? That is, does the fact that $R(U)$ and $N(U)$ are $T$-invariant imply that $U$ is invertible? That would then prove that $UT=TU$, correct?
– notadoctor
Nov 15 at 2:19
Cool problem. Where did you get this problem?
– mathnoob
Nov 15 at 2:23
2
@justadampaul No. Consider $U=pmatrix{1\ &2\ &&0}, T=pmatrix{0&1\ 1&0\ &&1}$. Then $R(U)={(ast,ast,0)^T}$ and $N(U)={(0,0,ast)^T}$ are $T$-invariant, but $U$ is singular and $UTne TU$.
– user1551
Nov 15 at 2:33
1
It doesn't have to be non-invertible. user1551's first comment suggests the opposite. Consider two rotations in 3D; both are invertible and have trivial nullspace and range, but rotations are not commutative in general.
– mr_e_man
Nov 15 at 2:50
2
2
If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$.
– user1551
Nov 15 at 2:14
If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$.
– user1551
Nov 15 at 2:14
Is the converse of this true? That is, does the fact that $R(U)$ and $N(U)$ are $T$-invariant imply that $U$ is invertible? That would then prove that $UT=TU$, correct?
– notadoctor
Nov 15 at 2:19
Is the converse of this true? That is, does the fact that $R(U)$ and $N(U)$ are $T$-invariant imply that $U$ is invertible? That would then prove that $UT=TU$, correct?
– notadoctor
Nov 15 at 2:19
Cool problem. Where did you get this problem?
– mathnoob
Nov 15 at 2:23
Cool problem. Where did you get this problem?
– mathnoob
Nov 15 at 2:23
2
2
@justadampaul No. Consider $U=pmatrix{1\ &2\ &&0}, T=pmatrix{0&1\ 1&0\ &&1}$. Then $R(U)={(ast,ast,0)^T}$ and $N(U)={(0,0,ast)^T}$ are $T$-invariant, but $U$ is singular and $UTne TU$.
– user1551
Nov 15 at 2:33
@justadampaul No. Consider $U=pmatrix{1\ &2\ &&0}, T=pmatrix{0&1\ 1&0\ &&1}$. Then $R(U)={(ast,ast,0)^T}$ and $N(U)={(0,0,ast)^T}$ are $T$-invariant, but $U$ is singular and $UTne TU$.
– user1551
Nov 15 at 2:33
1
1
It doesn't have to be non-invertible. user1551's first comment suggests the opposite. Consider two rotations in 3D; both are invertible and have trivial nullspace and range, but rotations are not commutative in general.
– mr_e_man
Nov 15 at 2:50
It doesn't have to be non-invertible. user1551's first comment suggests the opposite. Consider two rotations in 3D; both are invertible and have trivial nullspace and range, but rotations are not commutative in general.
– mr_e_man
Nov 15 at 2:50
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Via user mr_e_man in the comments above: No. Rotations in $R^3$ are not generally commutative, but two rotations $T$ and $U$ have nullspaces and ranges which are equal (and therefore $T$-invariant).
answered Nov 15 at 19:19
notadoctor
867
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Via user mr_e_man in the comments above: No. Rotations in $R^3$ are not generally commutative, but two rotations $T$ and $U$ have nullspaces and ranges which are equal (and therefore $T$-invariant).
answered Nov 15 at 19:19
notadoctor
867
add a comment |
up vote
0
down vote
accepted
Via user mr_e_man in the comments above: No. Rotations in $R^3$ are not generally commutative, but two rotations $T$ and $U$ have nullspaces and ranges which are equal (and therefore $T$-invariant).
answered Nov 15 at 19:19
notadoctor
867
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Via user mr_e_man in the comments above: No. Rotations in $R^3$ are not generally commutative, but two rotations $T$ and $U$ have nullspaces and ranges which are equal (and therefore $T$-invariant).
answered Nov 15 at 19:19
notadoctor
867
Via user mr_e_man in the comments above: No. Rotations in $R^3$ are not generally commutative, but two rotations $T$ and $U$ have nullspaces and ranges which are equal (and therefore $T$-invariant).
answered Nov 15 at 19:19
notadoctor
867
answered Nov 15 at 19:19
notadoctor
867
answered Nov 15 at 19:19
notadoctor
867
answered Nov 15 at 19:19
notadoctor
867
867
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999092%2fdoes-transformation-invariance-of-the-range-and-null-space-imply-commutativity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$.
– user1551
Nov 15 at 2:14
Is the converse of this true? That is, does the fact that $R(U)$ and $N(U)$ are $T$-invariant imply that $U$ is invertible? That would then prove that $UT=TU$, correct?
– notadoctor
Nov 15 at 2:19
Cool problem. Where did you get this problem?
– mathnoob
Nov 15 at 2:23
2
@justadampaul No. Consider $U=pmatrix{1\ &2\ &&0}, T=pmatrix{0&1\ 1&0\ &&1}$. Then $R(U)={(ast,ast,0)^T}$ and $N(U)={(0,0,ast)^T}$ are $T$-invariant, but $U$ is singular and $UTne TU$.
– user1551
Nov 15 at 2:33
1
It doesn't have to be non-invertible. user1551's first comment suggests the opposite. Consider two rotations in 3D; both are invertible and have trivial nullspace and range, but rotations are not commutative in general.
– mr_e_man
Nov 15 at 2:50