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Existence and Characterization of spaces of singular matrices
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I'm reading baby Rudin, and I know that the set of invertible matrices forms an open set (under the topology induced by the metric induced by the norm $||A|| = sup$ where $x in mathbbR^n$). I also know that this set is a disconnected union of two connected components: matrices with positive determinant and matrices with negative determinant.
My topological intuition says the set of singular matrices is an $n$-manifold, and my guess is that $n = 2$. I am fairly confident that this is true for 2x2 matrices (because it can be identified with $mathbbR^4$; I know they have different norms) but it quickly becomes difficult to picture this (nine dimensions are a bit too much for me). I would be very interested to see a proof, or a sketch of one, on this conjecture, but I unfortunately don't have a great deal of experience with manifolds or local homeomorphisms. Thanks!
real-analysis matrices manifolds
asked Aug 2 at 3:26
Noah Ankney
261
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up vote
1
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I'm reading baby Rudin, and I know that the set of invertible matrices forms an open set (under the topology induced by the metric induced by the norm $||A|| = sup$ where $x in mathbbR^n$). I also know that this set is a disconnected union of two connected components: matrices with positive determinant and matrices with negative determinant.
My topological intuition says the set of singular matrices is an $n$-manifold, and my guess is that $n = 2$. I am fairly confident that this is true for 2x2 matrices (because it can be identified with $mathbbR^4$; I know they have different norms) but it quickly becomes difficult to picture this (nine dimensions are a bit too much for me). I would be very interested to see a proof, or a sketch of one, on this conjecture, but I unfortunately don't have a great deal of experience with manifolds or local homeomorphisms. Thanks!
real-analysis matrices manifolds
asked Aug 2 at 3:26
Noah Ankney
261
$(a,b,c,d)inmathbbR^4: ad-bc=0$ is singular at the origin $a=b=c=d=0$.
– spiralstotheleft
Aug 2 at 3:42
One can test if $S=(a,b,c,d)inmathbbR^4: ad-bc=0$ is a submanifold using the Jacobian criterion. The Jacobian of $ad-bc$ is $(d,-c,-b,a)$, which has rank $4-(4-1)=1$ at all points of $S$, except for the origin $a=b=c=d=0$. Therefore, $S$ is not a smooth submanifold of $mathbbR^4$. Now, this criterion doesn't tell you if it is a topological manifold or not. For example, $(x,y)inmathbbR^2: y^2-x^3=0$ doesn't satisfy the Jacobian criterion at $(0,0)$, but it is a topological manifold anyways. You can parameterize it with the chart $tmapsto (t^2,t^3)$.
– spiralstotheleft
Aug 2 at 13:45
The question if it is a topological manifold has been asked here before but no one has answered.
– spiralstotheleft
Aug 2 at 13:46
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading baby Rudin, and I know that the set of invertible matrices forms an open set (under the topology induced by the metric induced by the norm $||A|| = sup$ where $x in mathbbR^n$). I also know that this set is a disconnected union of two connected components: matrices with positive determinant and matrices with negative determinant.
My topological intuition says the set of singular matrices is an $n$-manifold, and my guess is that $n = 2$. I am fairly confident that this is true for 2x2 matrices (because it can be identified with $mathbbR^4$; I know they have different norms) but it quickly becomes difficult to picture this (nine dimensions are a bit too much for me). I would be very interested to see a proof, or a sketch of one, on this conjecture, but I unfortunately don't have a great deal of experience with manifolds or local homeomorphisms. Thanks!
real-analysis matrices manifolds
asked Aug 2 at 3:26
Noah Ankney
261
I'm reading baby Rudin, and I know that the set of invertible matrices forms an open set (under the topology induced by the metric induced by the norm $||A|| = sup$ where $x in mathbbR^n$). I also know that this set is a disconnected union of two connected components: matrices with positive determinant and matrices with negative determinant.
My topological intuition says the set of singular matrices is an $n$-manifold, and my guess is that $n = 2$. I am fairly confident that this is true for 2x2 matrices (because it can be identified with $mathbbR^4$; I know they have different norms) but it quickly becomes difficult to picture this (nine dimensions are a bit too much for me). I would be very interested to see a proof, or a sketch of one, on this conjecture, but I unfortunately don't have a great deal of experience with manifolds or local homeomorphisms. Thanks!
real-analysis matrices manifolds
asked Aug 2 at 3:26
Noah Ankney
261
asked Aug 2 at 3:26
Noah Ankney
261
asked Aug 2 at 3:26
Noah Ankney
261
asked Aug 2 at 3:26
Noah Ankney
261
261
$(a,b,c,d)inmathbbR^4: ad-bc=0$ is singular at the origin $a=b=c=d=0$.
– spiralstotheleft
Aug 2 at 3:42
One can test if $S=(a,b,c,d)inmathbbR^4: ad-bc=0$ is a submanifold using the Jacobian criterion. The Jacobian of $ad-bc$ is $(d,-c,-b,a)$, which has rank $4-(4-1)=1$ at all points of $S$, except for the origin $a=b=c=d=0$. Therefore, $S$ is not a smooth submanifold of $mathbbR^4$. Now, this criterion doesn't tell you if it is a topological manifold or not. For example, $(x,y)inmathbbR^2: y^2-x^3=0$ doesn't satisfy the Jacobian criterion at $(0,0)$, but it is a topological manifold anyways. You can parameterize it with the chart $tmapsto (t^2,t^3)$.
– spiralstotheleft
Aug 2 at 13:45
The question if it is a topological manifold has been asked here before but no one has answered.
– spiralstotheleft
Aug 2 at 13:46
add a comment |Â
$(a,b,c,d)inmathbbR^4: ad-bc=0$ is singular at the origin $a=b=c=d=0$.
– spiralstotheleft
Aug 2 at 3:42
One can test if $S=(a,b,c,d)inmathbbR^4: ad-bc=0$ is a submanifold using the Jacobian criterion. The Jacobian of $ad-bc$ is $(d,-c,-b,a)$, which has rank $4-(4-1)=1$ at all points of $S$, except for the origin $a=b=c=d=0$. Therefore, $S$ is not a smooth submanifold of $mathbbR^4$. Now, this criterion doesn't tell you if it is a topological manifold or not. For example, $(x,y)inmathbbR^2: y^2-x^3=0$ doesn't satisfy the Jacobian criterion at $(0,0)$, but it is a topological manifold anyways. You can parameterize it with the chart $tmapsto (t^2,t^3)$.
– spiralstotheleft
Aug 2 at 13:45
The question if it is a topological manifold has been asked here before but no one has answered.
– spiralstotheleft
Aug 2 at 13:46
$(a,b,c,d)inmathbbR^4: ad-bc=0$ is singular at the origin $a=b=c=d=0$.
– spiralstotheleft
Aug 2 at 3:42
$(a,b,c,d)inmathbbR^4: ad-bc=0$ is singular at the origin $a=b=c=d=0$.
– spiralstotheleft
Aug 2 at 3:42
One can test if $S=(a,b,c,d)inmathbbR^4: ad-bc=0$ is a submanifold using the Jacobian criterion. The Jacobian of $ad-bc$ is $(d,-c,-b,a)$, which has rank $4-(4-1)=1$ at all points of $S$, except for the origin $a=b=c=d=0$. Therefore, $S$ is not a smooth submanifold of $mathbbR^4$. Now, this criterion doesn't tell you if it is a topological manifold or not. For example, $(x,y)inmathbbR^2: y^2-x^3=0$ doesn't satisfy the Jacobian criterion at $(0,0)$, but it is a topological manifold anyways. You can parameterize it with the chart $tmapsto (t^2,t^3)$.
– spiralstotheleft
Aug 2 at 13:45
One can test if $S=(a,b,c,d)inmathbbR^4: ad-bc=0$ is a submanifold using the Jacobian criterion. The Jacobian of $ad-bc$ is $(d,-c,-b,a)$, which has rank $4-(4-1)=1$ at all points of $S$, except for the origin $a=b=c=d=0$. Therefore, $S$ is not a smooth submanifold of $mathbbR^4$. Now, this criterion doesn't tell you if it is a topological manifold or not. For example, $(x,y)inmathbbR^2: y^2-x^3=0$ doesn't satisfy the Jacobian criterion at $(0,0)$, but it is a topological manifold anyways. You can parameterize it with the chart $tmapsto (t^2,t^3)$.
– spiralstotheleft
Aug 2 at 13:45
The question if it is a topological manifold has been asked here before but no one has answered.
– spiralstotheleft
Aug 2 at 13:46
The question if it is a topological manifold has been asked here before but no one has answered.
– spiralstotheleft
Aug 2 at 13:46
add a comment |Â
1 Answer
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No, it's not true. A square matrix is singular iff its determinant is $0$. That means the singular matrices form a variety of dimension $n^2-1$ in the $n^2$-dimensional space of $n times n$ matrices. But it's not a manifold.
Thus in the case $n=2$, the cross-section $d=0$ of the set of singular matrices $pmatrixa & bcr c & dcr$ is the union of the two planes $b=0$ and $c=0$
intersecting at right angles.
answered Aug 2 at 3:48
Robert Israel
303k22200440
1
A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4.
– Lorenzo
Aug 2 at 5:37
(I like your answer I just think that point is potentially misleading.)
– Lorenzo
Aug 2 at 5:43
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
No, it's not true. A square matrix is singular iff its determinant is $0$. That means the singular matrices form a variety of dimension $n^2-1$ in the $n^2$-dimensional space of $n times n$ matrices. But it's not a manifold.
Thus in the case $n=2$, the cross-section $d=0$ of the set of singular matrices $pmatrixa & bcr c & dcr$ is the union of the two planes $b=0$ and $c=0$
intersecting at right angles.
answered Aug 2 at 3:48
Robert Israel
303k22200440
1
A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4.
– Lorenzo
Aug 2 at 5:37
(I like your answer I just think that point is potentially misleading.)
– Lorenzo
Aug 2 at 5:43
add a comment |Â
up vote
3
down vote
No, it's not true. A square matrix is singular iff its determinant is $0$. That means the singular matrices form a variety of dimension $n^2-1$ in the $n^2$-dimensional space of $n times n$ matrices. But it's not a manifold.
Thus in the case $n=2$, the cross-section $d=0$ of the set of singular matrices $pmatrixa & bcr c & dcr$ is the union of the two planes $b=0$ and $c=0$
intersecting at right angles.
answered Aug 2 at 3:48
Robert Israel
303k22200440
1
A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4.
– Lorenzo
Aug 2 at 5:37
(I like your answer I just think that point is potentially misleading.)
– Lorenzo
Aug 2 at 5:43
add a comment |Â
up vote
3
down vote
up vote
3
down vote
No, it's not true. A square matrix is singular iff its determinant is $0$. That means the singular matrices form a variety of dimension $n^2-1$ in the $n^2$-dimensional space of $n times n$ matrices. But it's not a manifold.
Thus in the case $n=2$, the cross-section $d=0$ of the set of singular matrices $pmatrixa & bcr c & dcr$ is the union of the two planes $b=0$ and $c=0$
intersecting at right angles.
answered Aug 2 at 3:48
Robert Israel
303k22200440
No, it's not true. A square matrix is singular iff its determinant is $0$. That means the singular matrices form a variety of dimension $n^2-1$ in the $n^2$-dimensional space of $n times n$ matrices. But it's not a manifold.
Thus in the case $n=2$, the cross-section $d=0$ of the set of singular matrices $pmatrixa & bcr c & dcr$ is the union of the two planes $b=0$ and $c=0$
intersecting at right angles.
answered Aug 2 at 3:48
Robert Israel
303k22200440
answered Aug 2 at 3:48
Robert Israel
303k22200440
answered Aug 2 at 3:48
Robert Israel
303k22200440
answered Aug 2 at 3:48
Robert Israel
303k22200440
303k22200440
1
A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4.
– Lorenzo
Aug 2 at 5:37
(I like your answer I just think that point is potentially misleading.)
– Lorenzo
Aug 2 at 5:43
add a comment |Â
1
A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4.
– Lorenzo
Aug 2 at 5:37
(I like your answer I just think that point is potentially misleading.)
– Lorenzo
Aug 2 at 5:43
1
1
A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4.
– Lorenzo
Aug 2 at 5:37
A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4.
– Lorenzo
Aug 2 at 5:37
(I like your answer I just think that point is potentially misleading.)
– Lorenzo
Aug 2 at 5:43
(I like your answer I just think that point is potentially misleading.)
– Lorenzo
Aug 2 at 5:43
add a comment |Â
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$(a,b,c,d)inmathbbR^4: ad-bc=0$ is singular at the origin $a=b=c=d=0$.
– spiralstotheleft
Aug 2 at 3:42
One can test if $S=(a,b,c,d)inmathbbR^4: ad-bc=0$ is a submanifold using the Jacobian criterion. The Jacobian of $ad-bc$ is $(d,-c,-b,a)$, which has rank $4-(4-1)=1$ at all points of $S$, except for the origin $a=b=c=d=0$. Therefore, $S$ is not a smooth submanifold of $mathbbR^4$. Now, this criterion doesn't tell you if it is a topological manifold or not. For example, $(x,y)inmathbbR^2: y^2-x^3=0$ doesn't satisfy the Jacobian criterion at $(0,0)$, but it is a topological manifold anyways. You can parameterize it with the chart $tmapsto (t^2,t^3)$.
– spiralstotheleft
Aug 2 at 13:45
The question if it is a topological manifold has been asked here before but no one has answered.
– spiralstotheleft
Aug 2 at 13:46