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When does $S cap GL_4(mathbb R)$ have precisely two connected components where $S subset M_4(mathbb R)$ is a linear subspace?
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This is related to the question How many connected components for the intersection $S cap GL_n(mathbb R)$ where $S subset M_n(mathbb R)$ is a linear subspace? I asked. There is a nice example in the answer to show the intersection does not need to have two connected components as $GL_n(mathbb R)$. The comment below by Travis is also very helpful.
Let $S subset M_n(mathbb R^n)$ be a linear subspace. What property should $S$ process such that $S cap GL_n(mathbb R)$ has two connected components?
I realized the question I asked before (crossed out above) might be too general to answer. Since it has not been answered, let me ask the specific question I am considering.
The question is now cross-posted here at MO.
Let $A in M_4(mathbb R)$ and $A = (e_2, x, e_4, y)$ where $e_2, e_4$ are standard basis in $mathbb R^4$ and $x,y$ are undetermined variables. Let $phi, psi: M_4(mathbb R) to mathbb R^4$ be linear maps defined by
beginalign*
&phi: B mapsto (AB-BA) e_1, \
&psi: B mapsto (AB-BA)e_3.
endalign*
The subspace $S$ I am interested in is the intersection of kernels of the two linear maps, i.e., $S :=textker(phi) cap textkerpsi$. In other words, the elements in $S cap GL_4(mathbb R)$ would preserve the structure of first and third columns of $A$ by conjugation, i.e., $(B^-1AB) e_1 = e_2, (B^-1AB)e_3 = e_4$ for $B in S cap GL_4(mathbb R)$. I would like to determine:
- whether there exists $A$
with eigenvalues all lying on the left open half plane of $mathbb C$, i.e., with negative real parts( we can freely choose $x, y$) such that $S cap GL_4(mathbb R)$ has precisely two connected components or precisely one component.
- If there exists $A$, such that $V^-1 A V: V in S cap GL_4(mathbb R)$ is connected.
Edit 1: If $S cap GL_4(mathbb R)$ has precisely two connected components, I guess they should be $S cap GL_4(mathbb R)_+$ and $S cap GL_4(mathbb R)_-$. So if $V in S$ and $det(V) > 0$, $V$ should be path-connected with $I$. It is not hard to check the condition implies $V = (v_1, Av_1, v_3, Av_3)$. Since $e_2 = Ae_1, e_4 = Ae_3$, the question is can we continuously change $v_1, v_3$ to $e_1, e_3$ such that $(v_1, Av_1, v_3, Av_3)$ stay linearly independent during the process.
Edit 2: If the intersection only have one component, then the $2^textnd$ question is immediate. Or if $A$ has two components but with a real eigenvalue, then $2$ should hold too. However, it is possible $2$ can be solve directly which I could not see.
Edit 3: I crossed out the restrictions I put on $A$ although I feel this should not matter too much. The second question is newly added which is actually my end question. Before I had a feeling there should be some "special" $A$ such that the intersection would only give $1$ or $2$ components. As mentioned above, it's highly possible we can directly attack the second question.
linear-algebra abstract-algebra general-topology matrices connectedness
asked Jul 22 at 0:37
user9527
925525
add a comment |Â
up vote
9
down vote
favorite
This is related to the question How many connected components for the intersection $S cap GL_n(mathbb R)$ where $S subset M_n(mathbb R)$ is a linear subspace? I asked. There is a nice example in the answer to show the intersection does not need to have two connected components as $GL_n(mathbb R)$. The comment below by Travis is also very helpful.
Let $S subset M_n(mathbb R^n)$ be a linear subspace. What property should $S$ process such that $S cap GL_n(mathbb R)$ has two connected components?
I realized the question I asked before (crossed out above) might be too general to answer. Since it has not been answered, let me ask the specific question I am considering.
The question is now cross-posted here at MO.
Let $A in M_4(mathbb R)$ and $A = (e_2, x, e_4, y)$ where $e_2, e_4$ are standard basis in $mathbb R^4$ and $x,y$ are undetermined variables. Let $phi, psi: M_4(mathbb R) to mathbb R^4$ be linear maps defined by
beginalign*
&phi: B mapsto (AB-BA) e_1, \
&psi: B mapsto (AB-BA)e_3.
endalign*
The subspace $S$ I am interested in is the intersection of kernels of the two linear maps, i.e., $S :=textker(phi) cap textkerpsi$. In other words, the elements in $S cap GL_4(mathbb R)$ would preserve the structure of first and third columns of $A$ by conjugation, i.e., $(B^-1AB) e_1 = e_2, (B^-1AB)e_3 = e_4$ for $B in S cap GL_4(mathbb R)$. I would like to determine:
- whether there exists $A$
with eigenvalues all lying on the left open half plane of $mathbb C$, i.e., with negative real parts( we can freely choose $x, y$) such that $S cap GL_4(mathbb R)$ has precisely two connected components or precisely one component.
- If there exists $A$, such that $V^-1 A V: V in S cap GL_4(mathbb R)$ is connected.
Edit 1: If $S cap GL_4(mathbb R)$ has precisely two connected components, I guess they should be $S cap GL_4(mathbb R)_+$ and $S cap GL_4(mathbb R)_-$. So if $V in S$ and $det(V) > 0$, $V$ should be path-connected with $I$. It is not hard to check the condition implies $V = (v_1, Av_1, v_3, Av_3)$. Since $e_2 = Ae_1, e_4 = Ae_3$, the question is can we continuously change $v_1, v_3$ to $e_1, e_3$ such that $(v_1, Av_1, v_3, Av_3)$ stay linearly independent during the process.
Edit 2: If the intersection only have one component, then the $2^textnd$ question is immediate. Or if $A$ has two components but with a real eigenvalue, then $2$ should hold too. However, it is possible $2$ can be solve directly which I could not see.
Edit 3: I crossed out the restrictions I put on $A$ although I feel this should not matter too much. The second question is newly added which is actually my end question. Before I had a feeling there should be some "special" $A$ such that the intersection would only give $1$ or $2$ components. As mentioned above, it's highly possible we can directly attack the second question.
linear-algebra abstract-algebra general-topology matrices connectedness
asked Jul 22 at 0:37
user9527
925525
To be clear, do you mean precisely two components?
– Travis
Jul 22 at 0:52
I didn't consider it would have one component. Actually I want to see when it has precisely $1$ component or precisely $2$ components.
– user9527
Jul 22 at 0:59
2
For $n$ odd, it's not possible to have just one component: If $A$ is in $S cap operatornameGL_n(Bbb R)$, then so is $-A$, but $det (-A) = - det A$, so $A, -A$ lie in different components of $GL_n(Bbb R)$. For $n$ even, it is possible: Consider $S := leftpmatrixa&-b\b&a : a, b, in Bbb Rright$. Then $det pmatrixa&-b\b&a = a^2 + b^2$, so $S cap operatornameGL_n(Bbb R) = S - 0$, which is connected.
– Travis
Jul 22 at 1:08
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
This is related to the question How many connected components for the intersection $S cap GL_n(mathbb R)$ where $S subset M_n(mathbb R)$ is a linear subspace? I asked. There is a nice example in the answer to show the intersection does not need to have two connected components as $GL_n(mathbb R)$. The comment below by Travis is also very helpful.
Let $S subset M_n(mathbb R^n)$ be a linear subspace. What property should $S$ process such that $S cap GL_n(mathbb R)$ has two connected components?
I realized the question I asked before (crossed out above) might be too general to answer. Since it has not been answered, let me ask the specific question I am considering.
The question is now cross-posted here at MO.
Let $A in M_4(mathbb R)$ and $A = (e_2, x, e_4, y)$ where $e_2, e_4$ are standard basis in $mathbb R^4$ and $x,y$ are undetermined variables. Let $phi, psi: M_4(mathbb R) to mathbb R^4$ be linear maps defined by
beginalign*
&phi: B mapsto (AB-BA) e_1, \
&psi: B mapsto (AB-BA)e_3.
endalign*
The subspace $S$ I am interested in is the intersection of kernels of the two linear maps, i.e., $S :=textker(phi) cap textkerpsi$. In other words, the elements in $S cap GL_4(mathbb R)$ would preserve the structure of first and third columns of $A$ by conjugation, i.e., $(B^-1AB) e_1 = e_2, (B^-1AB)e_3 = e_4$ for $B in S cap GL_4(mathbb R)$. I would like to determine:
- whether there exists $A$
with eigenvalues all lying on the left open half plane of $mathbb C$, i.e., with negative real parts( we can freely choose $x, y$) such that $S cap GL_4(mathbb R)$ has precisely two connected components or precisely one component.
- If there exists $A$, such that $V^-1 A V: V in S cap GL_4(mathbb R)$ is connected.
Edit 1: If $S cap GL_4(mathbb R)$ has precisely two connected components, I guess they should be $S cap GL_4(mathbb R)_+$ and $S cap GL_4(mathbb R)_-$. So if $V in S$ and $det(V) > 0$, $V$ should be path-connected with $I$. It is not hard to check the condition implies $V = (v_1, Av_1, v_3, Av_3)$. Since $e_2 = Ae_1, e_4 = Ae_3$, the question is can we continuously change $v_1, v_3$ to $e_1, e_3$ such that $(v_1, Av_1, v_3, Av_3)$ stay linearly independent during the process.
Edit 2: If the intersection only have one component, then the $2^textnd$ question is immediate. Or if $A$ has two components but with a real eigenvalue, then $2$ should hold too. However, it is possible $2$ can be solve directly which I could not see.
Edit 3: I crossed out the restrictions I put on $A$ although I feel this should not matter too much. The second question is newly added which is actually my end question. Before I had a feeling there should be some "special" $A$ such that the intersection would only give $1$ or $2$ components. As mentioned above, it's highly possible we can directly attack the second question.
linear-algebra abstract-algebra general-topology matrices connectedness
asked Jul 22 at 0:37
user9527
925525
This is related to the question How many connected components for the intersection $S cap GL_n(mathbb R)$ where $S subset M_n(mathbb R)$ is a linear subspace? I asked. There is a nice example in the answer to show the intersection does not need to have two connected components as $GL_n(mathbb R)$. The comment below by Travis is also very helpful.
Let $S subset M_n(mathbb R^n)$ be a linear subspace. What property should $S$ process such that $S cap GL_n(mathbb R)$ has two connected components?
I realized the question I asked before (crossed out above) might be too general to answer. Since it has not been answered, let me ask the specific question I am considering.
The question is now cross-posted here at MO.
Let $A in M_4(mathbb R)$ and $A = (e_2, x, e_4, y)$ where $e_2, e_4$ are standard basis in $mathbb R^4$ and $x,y$ are undetermined variables. Let $phi, psi: M_4(mathbb R) to mathbb R^4$ be linear maps defined by
beginalign*
&phi: B mapsto (AB-BA) e_1, \
&psi: B mapsto (AB-BA)e_3.
endalign*
The subspace $S$ I am interested in is the intersection of kernels of the two linear maps, i.e., $S :=textker(phi) cap textkerpsi$. In other words, the elements in $S cap GL_4(mathbb R)$ would preserve the structure of first and third columns of $A$ by conjugation, i.e., $(B^-1AB) e_1 = e_2, (B^-1AB)e_3 = e_4$ for $B in S cap GL_4(mathbb R)$. I would like to determine:
- whether there exists $A$
with eigenvalues all lying on the left open half plane of $mathbb C$, i.e., with negative real parts( we can freely choose $x, y$) such that $S cap GL_4(mathbb R)$ has precisely two connected components or precisely one component.
- If there exists $A$, such that $V^-1 A V: V in S cap GL_4(mathbb R)$ is connected.
Edit 1: If $S cap GL_4(mathbb R)$ has precisely two connected components, I guess they should be $S cap GL_4(mathbb R)_+$ and $S cap GL_4(mathbb R)_-$. So if $V in S$ and $det(V) > 0$, $V$ should be path-connected with $I$. It is not hard to check the condition implies $V = (v_1, Av_1, v_3, Av_3)$. Since $e_2 = Ae_1, e_4 = Ae_3$, the question is can we continuously change $v_1, v_3$ to $e_1, e_3$ such that $(v_1, Av_1, v_3, Av_3)$ stay linearly independent during the process.
Edit 2: If the intersection only have one component, then the $2^textnd$ question is immediate. Or if $A$ has two components but with a real eigenvalue, then $2$ should hold too. However, it is possible $2$ can be solve directly which I could not see.
Edit 3: I crossed out the restrictions I put on $A$ although I feel this should not matter too much. The second question is newly added which is actually my end question. Before I had a feeling there should be some "special" $A$ such that the intersection would only give $1$ or $2$ components. As mentioned above, it's highly possible we can directly attack the second question.
linear-algebra abstract-algebra general-topology matrices connectedness
asked Jul 22 at 0:37
user9527
925525
edited Aug 1 at 6:31
asked Jul 22 at 0:37
user9527
925525
asked Jul 22 at 0:37
user9527
925525
asked Jul 22 at 0:37
user9527
925525
925525
To be clear, do you mean precisely two components?
– Travis
Jul 22 at 0:52
I didn't consider it would have one component. Actually I want to see when it has precisely $1$ component or precisely $2$ components.
– user9527
Jul 22 at 0:59
2
For $n$ odd, it's not possible to have just one component: If $A$ is in $S cap operatornameGL_n(Bbb R)$, then so is $-A$, but $det (-A) = - det A$, so $A, -A$ lie in different components of $GL_n(Bbb R)$. For $n$ even, it is possible: Consider $S := leftpmatrixa&-b\b&a : a, b, in Bbb Rright$. Then $det pmatrixa&-b\b&a = a^2 + b^2$, so $S cap operatornameGL_n(Bbb R) = S - 0$, which is connected.
– Travis
Jul 22 at 1:08
add a comment |Â
To be clear, do you mean precisely two components?
– Travis
Jul 22 at 0:52
I didn't consider it would have one component. Actually I want to see when it has precisely $1$ component or precisely $2$ components.
– user9527
Jul 22 at 0:59
2
For $n$ odd, it's not possible to have just one component: If $A$ is in $S cap operatornameGL_n(Bbb R)$, then so is $-A$, but $det (-A) = - det A$, so $A, -A$ lie in different components of $GL_n(Bbb R)$. For $n$ even, it is possible: Consider $S := leftpmatrixa&-b\b&a : a, b, in Bbb Rright$. Then $det pmatrixa&-b\b&a = a^2 + b^2$, so $S cap operatornameGL_n(Bbb R) = S - 0$, which is connected.
– Travis
Jul 22 at 1:08
To be clear, do you mean precisely two components?
– Travis
Jul 22 at 0:52
To be clear, do you mean precisely two components?
– Travis
Jul 22 at 0:52
I didn't consider it would have one component. Actually I want to see when it has precisely $1$ component or precisely $2$ components.
– user9527
Jul 22 at 0:59
I didn't consider it would have one component. Actually I want to see when it has precisely $1$ component or precisely $2$ components.
– user9527
Jul 22 at 0:59
2
2
For $n$ odd, it's not possible to have just one component: If $A$ is in $S cap operatornameGL_n(Bbb R)$, then so is $-A$, but $det (-A) = - det A$, so $A, -A$ lie in different components of $GL_n(Bbb R)$. For $n$ even, it is possible: Consider $S := leftpmatrixa&-b\b&a : a, b, in Bbb Rright$. Then $det pmatrixa&-b\b&a = a^2 + b^2$, so $S cap operatornameGL_n(Bbb R) = S - 0$, which is connected.
– Travis
Jul 22 at 1:08
For $n$ odd, it's not possible to have just one component: If $A$ is in $S cap operatornameGL_n(Bbb R)$, then so is $-A$, but $det (-A) = - det A$, so $A, -A$ lie in different components of $GL_n(Bbb R)$. For $n$ even, it is possible: Consider $S := leftpmatrixa&-b\b&a : a, b, in Bbb Rright$. Then $det pmatrixa&-b\b&a = a^2 + b^2$, so $S cap operatornameGL_n(Bbb R) = S - 0$, which is connected.
– Travis
Jul 22 at 1:08
add a comment |Â
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To be clear, do you mean precisely two components?
– Travis
Jul 22 at 0:52
I didn't consider it would have one component. Actually I want to see when it has precisely $1$ component or precisely $2$ components.
– user9527
Jul 22 at 0:59
2
For $n$ odd, it's not possible to have just one component: If $A$ is in $S cap operatornameGL_n(Bbb R)$, then so is $-A$, but $det (-A) = - det A$, so $A, -A$ lie in different components of $GL_n(Bbb R)$. For $n$ even, it is possible: Consider $S := leftpmatrixa&-b\b&a : a, b, in Bbb Rright$. Then $det pmatrixa&-b\b&a = a^2 + b^2$, so $S cap operatornameGL_n(Bbb R) = S - 0$, which is connected.
– Travis
Jul 22 at 1:08