Non singular matrices are a differentiable submanifold
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How to understand why the set of nonsigular matrices is a differentiable submanifold of the the of matrices (of size $n$).
I thought of introducing, given a nonsingular matrix $A$, the mapping $f:Xmapsto det(X)-det(A)$. But from this on I do not know how to proceed.
Also, how to determine the tangent space at a point $A$?
matrices differential-geometry smooth-manifolds
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How to understand why the set of nonsigular matrices is a differentiable submanifold of the the of matrices (of size $n$).
I thought of introducing, given a nonsingular matrix $A$, the mapping $f:Xmapsto det(X)-det(A)$. But from this on I do not know how to proceed.
Also, how to determine the tangent space at a point $A$?
matrices differential-geometry smooth-manifolds
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to understand why the set of nonsigular matrices is a differentiable submanifold of the the of matrices (of size $n$).
I thought of introducing, given a nonsingular matrix $A$, the mapping $f:Xmapsto det(X)-det(A)$. But from this on I do not know how to proceed.
Also, how to determine the tangent space at a point $A$?
matrices differential-geometry smooth-manifolds
How to understand why the set of nonsigular matrices is a differentiable submanifold of the the of matrices (of size $n$).
I thought of introducing, given a nonsingular matrix $A$, the mapping $f:Xmapsto det(X)-det(A)$. But from this on I do not know how to proceed.
Also, how to determine the tangent space at a point $A$?
matrices differential-geometry smooth-manifolds
matrices differential-geometry smooth-manifolds
asked Dec 2 at 13:47
hecho
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The idea is that the singular matrices in $M_{m times q}(Bbb R)$ form a closed subset, as they constitute precisely the subset $text{det}^{-1}({0})$, and $text{det}$ is a continuous function on $M_{m times q}(Bbb R)$. So the subset of nonsingular matrices is the open set $M_{m times q}(Bbb R) setminus text{det}^{-1}({0})$, and any open subset of a Euclidean space is a submanifold of the same dimension (in this case, $mq$). For any nonsingular matrix $mathbf{A}$, the tangent space to $mathbf{A}$ in this open set is the same as $T_{mathbf{A}} M_{m times q}(Bbb R)$, which can be identified with $Bbb R^{mq}$.
More generally, let $Sigma_k subset M_{m times q}(Bbb R)$ be the space of matrices of rank $k$. For any matrix $mathbf{A}$ in $Sigma_k$, there is a $(k times k)$-minor, let's call it $A$, in $mathbf{A}$ which is nonsingular, and let the rest of the elements be arranged in three matrices $B_{k times (q-k)}$, $C_{(m-k) times k}$ and $D_{(m-k)times(q-k)}$, where the arrangement is done by performing $k$ row operations $E_1$ and $k$ column operations $E_2$ so that in $E_2E_1 mathbf{A}$, the minor $A$ is positioned at the upper left corner of the matrix, and $B, C, D$ are the other blocks. Since nonsingular matrices form an open subset, there is a neighborhood $U$ of $mathbf{A}$ in $M_{m times q}(Bbb R)$ where for any $mathbf{A}_0 in U$, the $(k times k)$-minor situated in the same position that $A$ is situated in $mathbf{A}$ is also nonsingular. Therefore, the blocks $A, B, C, D$ of $mathbf{A}$ are globally defined for every matrix in the open neighborhood $U$ of $mathbf{A}$.
Let $mathbf{X} in U$ be arbitrary and let $A, B, C, D$ be the blocks of $mathbf{X}$ as defined above. Notice that
$$mathbf{X} cdot begin{pmatrix} I & -A^{-1}B \ mathsf{O} & I end{pmatrix} = begin{pmatrix} A & mathsf{O} \ C & D - CA^{-1}B end{pmatrix}$$
Since the matrix we multiplied with is invertible, rank has remained unchanged. If $D - CA^{-1}B$ was nonzero we would have found a row in the bottom $(m-k)$-half of the matrix with a nonzero entry at one of the last $(q-k)$-th positions. This would then be linearly independent from the first $k$ rows of the matrix (as their last $(q-k)$ entries are all zero), which are themselves linearly independent by hypothesis on the minor $A$. That makes $text{rank}(mathbf{X}) geq k+1$, absurd. Therefore $D - CA^{-1}B = mathsf{O}$ is a necessary condition for $text{rank}(mathbf{X}) = k$. Once this condition holds the matrix above will have $k$ columns of vectors in $Bbb R^m$ whose projection to the first $q$ coordinates are linearly independent, hence are themselves linearly independent.
So consider the map $F : U to M_{(m-k)(q-k)}(Bbb R)$ given by $F(mathbf{X})= D - CA^{-1}B$. It's an easy check that $F$ is a smooth submersion, so $F^{-1}(mathsf{O}) = U cap Sigma_k$ is a submanifold of codimension $(m-k)(q-k)$ in $U$, which proves that $Sigma_k$ is a submanifold of $M_{m times q}(Bbb R)$.
This is not a closed, but a locally closed submanifold, of $M_{m times q}(Bbb R)$, and the closure $overline{Sigma_k}$ is precisely the subset of rank $leq k$ matrices in $M_{m times q}(Bbb R)$. This can be seen by taking any matrix $mathbf{M}$ with $text{rank}(mathbf{M}) = l < k$, which implies the $q$ column vectors $v_1, cdots, v_q$ span a subspace of dimension $l$ in $Bbb R^q$. Without loss of generality suppose $v_1, cdots, v_l$ is a basis of this subspace. Then $text{Span}(v_{l+1}), cdots, text{Span}
(v_{l+k})$ belong to this $l$-dimensional subspace; by a small perturbation in $Bbb R^q$ we can make them transverse to this $l$-dimensional subspace as well as with themselves. Let $w_{l+1}, cdots, w_{l+k}$ be the modified vectors; the matrix formed by the column vectors $v_1, cdots, v_l, w_{l+1}, cdots, w_{l+k}, v_{l+k+1}, cdots, v_q$ is a perturbation of $mathbf{M}$ which has rank $k$ (Note: if $k > q$, do the same argument with the $m$ rows). However, rank can never go down, as transverse subspaces remain transverse under small perturbations.
Therefore, $Sigma_k subset M_{m times q}(Bbb R)$ are locally closed submanifolds such that $overline{Sigma_k} = Sigma_0 cup cdots cup Sigma_{k-1}$. This kind of iterative structure is called a stratification; indeed, $Sigma_k$'s constitute stratums for a Whitney stratification of $M_{m times q}(Bbb R)$. But this was a long digression.
(+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
– loup blanc
Dec 4 at 11:33
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It is an open subspace of a vector space, $Gl(n,mathbb{R})=det^{-1}(mathbb{R}-{0}$. Here the vector space is the space of $ntimes n$-matrices so it is a submaniflod with one chart which is the embedding map $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
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up vote
1
down vote
The idea is that the singular matrices in $M_{m times q}(Bbb R)$ form a closed subset, as they constitute precisely the subset $text{det}^{-1}({0})$, and $text{det}$ is a continuous function on $M_{m times q}(Bbb R)$. So the subset of nonsingular matrices is the open set $M_{m times q}(Bbb R) setminus text{det}^{-1}({0})$, and any open subset of a Euclidean space is a submanifold of the same dimension (in this case, $mq$). For any nonsingular matrix $mathbf{A}$, the tangent space to $mathbf{A}$ in this open set is the same as $T_{mathbf{A}} M_{m times q}(Bbb R)$, which can be identified with $Bbb R^{mq}$.
More generally, let $Sigma_k subset M_{m times q}(Bbb R)$ be the space of matrices of rank $k$. For any matrix $mathbf{A}$ in $Sigma_k$, there is a $(k times k)$-minor, let's call it $A$, in $mathbf{A}$ which is nonsingular, and let the rest of the elements be arranged in three matrices $B_{k times (q-k)}$, $C_{(m-k) times k}$ and $D_{(m-k)times(q-k)}$, where the arrangement is done by performing $k$ row operations $E_1$ and $k$ column operations $E_2$ so that in $E_2E_1 mathbf{A}$, the minor $A$ is positioned at the upper left corner of the matrix, and $B, C, D$ are the other blocks. Since nonsingular matrices form an open subset, there is a neighborhood $U$ of $mathbf{A}$ in $M_{m times q}(Bbb R)$ where for any $mathbf{A}_0 in U$, the $(k times k)$-minor situated in the same position that $A$ is situated in $mathbf{A}$ is also nonsingular. Therefore, the blocks $A, B, C, D$ of $mathbf{A}$ are globally defined for every matrix in the open neighborhood $U$ of $mathbf{A}$.
Let $mathbf{X} in U$ be arbitrary and let $A, B, C, D$ be the blocks of $mathbf{X}$ as defined above. Notice that
$$mathbf{X} cdot begin{pmatrix} I & -A^{-1}B \ mathsf{O} & I end{pmatrix} = begin{pmatrix} A & mathsf{O} \ C & D - CA^{-1}B end{pmatrix}$$
Since the matrix we multiplied with is invertible, rank has remained unchanged. If $D - CA^{-1}B$ was nonzero we would have found a row in the bottom $(m-k)$-half of the matrix with a nonzero entry at one of the last $(q-k)$-th positions. This would then be linearly independent from the first $k$ rows of the matrix (as their last $(q-k)$ entries are all zero), which are themselves linearly independent by hypothesis on the minor $A$. That makes $text{rank}(mathbf{X}) geq k+1$, absurd. Therefore $D - CA^{-1}B = mathsf{O}$ is a necessary condition for $text{rank}(mathbf{X}) = k$. Once this condition holds the matrix above will have $k$ columns of vectors in $Bbb R^m$ whose projection to the first $q$ coordinates are linearly independent, hence are themselves linearly independent.
So consider the map $F : U to M_{(m-k)(q-k)}(Bbb R)$ given by $F(mathbf{X})= D - CA^{-1}B$. It's an easy check that $F$ is a smooth submersion, so $F^{-1}(mathsf{O}) = U cap Sigma_k$ is a submanifold of codimension $(m-k)(q-k)$ in $U$, which proves that $Sigma_k$ is a submanifold of $M_{m times q}(Bbb R)$.
This is not a closed, but a locally closed submanifold, of $M_{m times q}(Bbb R)$, and the closure $overline{Sigma_k}$ is precisely the subset of rank $leq k$ matrices in $M_{m times q}(Bbb R)$. This can be seen by taking any matrix $mathbf{M}$ with $text{rank}(mathbf{M}) = l < k$, which implies the $q$ column vectors $v_1, cdots, v_q$ span a subspace of dimension $l$ in $Bbb R^q$. Without loss of generality suppose $v_1, cdots, v_l$ is a basis of this subspace. Then $text{Span}(v_{l+1}), cdots, text{Span}
(v_{l+k})$ belong to this $l$-dimensional subspace; by a small perturbation in $Bbb R^q$ we can make them transverse to this $l$-dimensional subspace as well as with themselves. Let $w_{l+1}, cdots, w_{l+k}$ be the modified vectors; the matrix formed by the column vectors $v_1, cdots, v_l, w_{l+1}, cdots, w_{l+k}, v_{l+k+1}, cdots, v_q$ is a perturbation of $mathbf{M}$ which has rank $k$ (Note: if $k > q$, do the same argument with the $m$ rows). However, rank can never go down, as transverse subspaces remain transverse under small perturbations.
Therefore, $Sigma_k subset M_{m times q}(Bbb R)$ are locally closed submanifolds such that $overline{Sigma_k} = Sigma_0 cup cdots cup Sigma_{k-1}$. This kind of iterative structure is called a stratification; indeed, $Sigma_k$'s constitute stratums for a Whitney stratification of $M_{m times q}(Bbb R)$. But this was a long digression.
(+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
– loup blanc
Dec 4 at 11:33
add a comment |
up vote
1
down vote
The idea is that the singular matrices in $M_{m times q}(Bbb R)$ form a closed subset, as they constitute precisely the subset $text{det}^{-1}({0})$, and $text{det}$ is a continuous function on $M_{m times q}(Bbb R)$. So the subset of nonsingular matrices is the open set $M_{m times q}(Bbb R) setminus text{det}^{-1}({0})$, and any open subset of a Euclidean space is a submanifold of the same dimension (in this case, $mq$). For any nonsingular matrix $mathbf{A}$, the tangent space to $mathbf{A}$ in this open set is the same as $T_{mathbf{A}} M_{m times q}(Bbb R)$, which can be identified with $Bbb R^{mq}$.
More generally, let $Sigma_k subset M_{m times q}(Bbb R)$ be the space of matrices of rank $k$. For any matrix $mathbf{A}$ in $Sigma_k$, there is a $(k times k)$-minor, let's call it $A$, in $mathbf{A}$ which is nonsingular, and let the rest of the elements be arranged in three matrices $B_{k times (q-k)}$, $C_{(m-k) times k}$ and $D_{(m-k)times(q-k)}$, where the arrangement is done by performing $k$ row operations $E_1$ and $k$ column operations $E_2$ so that in $E_2E_1 mathbf{A}$, the minor $A$ is positioned at the upper left corner of the matrix, and $B, C, D$ are the other blocks. Since nonsingular matrices form an open subset, there is a neighborhood $U$ of $mathbf{A}$ in $M_{m times q}(Bbb R)$ where for any $mathbf{A}_0 in U$, the $(k times k)$-minor situated in the same position that $A$ is situated in $mathbf{A}$ is also nonsingular. Therefore, the blocks $A, B, C, D$ of $mathbf{A}$ are globally defined for every matrix in the open neighborhood $U$ of $mathbf{A}$.
Let $mathbf{X} in U$ be arbitrary and let $A, B, C, D$ be the blocks of $mathbf{X}$ as defined above. Notice that
$$mathbf{X} cdot begin{pmatrix} I & -A^{-1}B \ mathsf{O} & I end{pmatrix} = begin{pmatrix} A & mathsf{O} \ C & D - CA^{-1}B end{pmatrix}$$
Since the matrix we multiplied with is invertible, rank has remained unchanged. If $D - CA^{-1}B$ was nonzero we would have found a row in the bottom $(m-k)$-half of the matrix with a nonzero entry at one of the last $(q-k)$-th positions. This would then be linearly independent from the first $k$ rows of the matrix (as their last $(q-k)$ entries are all zero), which are themselves linearly independent by hypothesis on the minor $A$. That makes $text{rank}(mathbf{X}) geq k+1$, absurd. Therefore $D - CA^{-1}B = mathsf{O}$ is a necessary condition for $text{rank}(mathbf{X}) = k$. Once this condition holds the matrix above will have $k$ columns of vectors in $Bbb R^m$ whose projection to the first $q$ coordinates are linearly independent, hence are themselves linearly independent.
So consider the map $F : U to M_{(m-k)(q-k)}(Bbb R)$ given by $F(mathbf{X})= D - CA^{-1}B$. It's an easy check that $F$ is a smooth submersion, so $F^{-1}(mathsf{O}) = U cap Sigma_k$ is a submanifold of codimension $(m-k)(q-k)$ in $U$, which proves that $Sigma_k$ is a submanifold of $M_{m times q}(Bbb R)$.
This is not a closed, but a locally closed submanifold, of $M_{m times q}(Bbb R)$, and the closure $overline{Sigma_k}$ is precisely the subset of rank $leq k$ matrices in $M_{m times q}(Bbb R)$. This can be seen by taking any matrix $mathbf{M}$ with $text{rank}(mathbf{M}) = l < k$, which implies the $q$ column vectors $v_1, cdots, v_q$ span a subspace of dimension $l$ in $Bbb R^q$. Without loss of generality suppose $v_1, cdots, v_l$ is a basis of this subspace. Then $text{Span}(v_{l+1}), cdots, text{Span}
(v_{l+k})$ belong to this $l$-dimensional subspace; by a small perturbation in $Bbb R^q$ we can make them transverse to this $l$-dimensional subspace as well as with themselves. Let $w_{l+1}, cdots, w_{l+k}$ be the modified vectors; the matrix formed by the column vectors $v_1, cdots, v_l, w_{l+1}, cdots, w_{l+k}, v_{l+k+1}, cdots, v_q$ is a perturbation of $mathbf{M}$ which has rank $k$ (Note: if $k > q$, do the same argument with the $m$ rows). However, rank can never go down, as transverse subspaces remain transverse under small perturbations.
Therefore, $Sigma_k subset M_{m times q}(Bbb R)$ are locally closed submanifolds such that $overline{Sigma_k} = Sigma_0 cup cdots cup Sigma_{k-1}$. This kind of iterative structure is called a stratification; indeed, $Sigma_k$'s constitute stratums for a Whitney stratification of $M_{m times q}(Bbb R)$. But this was a long digression.
(+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
– loup blanc
Dec 4 at 11:33
add a comment |
up vote
1
down vote
up vote
1
down vote
The idea is that the singular matrices in $M_{m times q}(Bbb R)$ form a closed subset, as they constitute precisely the subset $text{det}^{-1}({0})$, and $text{det}$ is a continuous function on $M_{m times q}(Bbb R)$. So the subset of nonsingular matrices is the open set $M_{m times q}(Bbb R) setminus text{det}^{-1}({0})$, and any open subset of a Euclidean space is a submanifold of the same dimension (in this case, $mq$). For any nonsingular matrix $mathbf{A}$, the tangent space to $mathbf{A}$ in this open set is the same as $T_{mathbf{A}} M_{m times q}(Bbb R)$, which can be identified with $Bbb R^{mq}$.
More generally, let $Sigma_k subset M_{m times q}(Bbb R)$ be the space of matrices of rank $k$. For any matrix $mathbf{A}$ in $Sigma_k$, there is a $(k times k)$-minor, let's call it $A$, in $mathbf{A}$ which is nonsingular, and let the rest of the elements be arranged in three matrices $B_{k times (q-k)}$, $C_{(m-k) times k}$ and $D_{(m-k)times(q-k)}$, where the arrangement is done by performing $k$ row operations $E_1$ and $k$ column operations $E_2$ so that in $E_2E_1 mathbf{A}$, the minor $A$ is positioned at the upper left corner of the matrix, and $B, C, D$ are the other blocks. Since nonsingular matrices form an open subset, there is a neighborhood $U$ of $mathbf{A}$ in $M_{m times q}(Bbb R)$ where for any $mathbf{A}_0 in U$, the $(k times k)$-minor situated in the same position that $A$ is situated in $mathbf{A}$ is also nonsingular. Therefore, the blocks $A, B, C, D$ of $mathbf{A}$ are globally defined for every matrix in the open neighborhood $U$ of $mathbf{A}$.
Let $mathbf{X} in U$ be arbitrary and let $A, B, C, D$ be the blocks of $mathbf{X}$ as defined above. Notice that
$$mathbf{X} cdot begin{pmatrix} I & -A^{-1}B \ mathsf{O} & I end{pmatrix} = begin{pmatrix} A & mathsf{O} \ C & D - CA^{-1}B end{pmatrix}$$
Since the matrix we multiplied with is invertible, rank has remained unchanged. If $D - CA^{-1}B$ was nonzero we would have found a row in the bottom $(m-k)$-half of the matrix with a nonzero entry at one of the last $(q-k)$-th positions. This would then be linearly independent from the first $k$ rows of the matrix (as their last $(q-k)$ entries are all zero), which are themselves linearly independent by hypothesis on the minor $A$. That makes $text{rank}(mathbf{X}) geq k+1$, absurd. Therefore $D - CA^{-1}B = mathsf{O}$ is a necessary condition for $text{rank}(mathbf{X}) = k$. Once this condition holds the matrix above will have $k$ columns of vectors in $Bbb R^m$ whose projection to the first $q$ coordinates are linearly independent, hence are themselves linearly independent.
So consider the map $F : U to M_{(m-k)(q-k)}(Bbb R)$ given by $F(mathbf{X})= D - CA^{-1}B$. It's an easy check that $F$ is a smooth submersion, so $F^{-1}(mathsf{O}) = U cap Sigma_k$ is a submanifold of codimension $(m-k)(q-k)$ in $U$, which proves that $Sigma_k$ is a submanifold of $M_{m times q}(Bbb R)$.
This is not a closed, but a locally closed submanifold, of $M_{m times q}(Bbb R)$, and the closure $overline{Sigma_k}$ is precisely the subset of rank $leq k$ matrices in $M_{m times q}(Bbb R)$. This can be seen by taking any matrix $mathbf{M}$ with $text{rank}(mathbf{M}) = l < k$, which implies the $q$ column vectors $v_1, cdots, v_q$ span a subspace of dimension $l$ in $Bbb R^q$. Without loss of generality suppose $v_1, cdots, v_l$ is a basis of this subspace. Then $text{Span}(v_{l+1}), cdots, text{Span}
(v_{l+k})$ belong to this $l$-dimensional subspace; by a small perturbation in $Bbb R^q$ we can make them transverse to this $l$-dimensional subspace as well as with themselves. Let $w_{l+1}, cdots, w_{l+k}$ be the modified vectors; the matrix formed by the column vectors $v_1, cdots, v_l, w_{l+1}, cdots, w_{l+k}, v_{l+k+1}, cdots, v_q$ is a perturbation of $mathbf{M}$ which has rank $k$ (Note: if $k > q$, do the same argument with the $m$ rows). However, rank can never go down, as transverse subspaces remain transverse under small perturbations.
Therefore, $Sigma_k subset M_{m times q}(Bbb R)$ are locally closed submanifolds such that $overline{Sigma_k} = Sigma_0 cup cdots cup Sigma_{k-1}$. This kind of iterative structure is called a stratification; indeed, $Sigma_k$'s constitute stratums for a Whitney stratification of $M_{m times q}(Bbb R)$. But this was a long digression.
The idea is that the singular matrices in $M_{m times q}(Bbb R)$ form a closed subset, as they constitute precisely the subset $text{det}^{-1}({0})$, and $text{det}$ is a continuous function on $M_{m times q}(Bbb R)$. So the subset of nonsingular matrices is the open set $M_{m times q}(Bbb R) setminus text{det}^{-1}({0})$, and any open subset of a Euclidean space is a submanifold of the same dimension (in this case, $mq$). For any nonsingular matrix $mathbf{A}$, the tangent space to $mathbf{A}$ in this open set is the same as $T_{mathbf{A}} M_{m times q}(Bbb R)$, which can be identified with $Bbb R^{mq}$.
More generally, let $Sigma_k subset M_{m times q}(Bbb R)$ be the space of matrices of rank $k$. For any matrix $mathbf{A}$ in $Sigma_k$, there is a $(k times k)$-minor, let's call it $A$, in $mathbf{A}$ which is nonsingular, and let the rest of the elements be arranged in three matrices $B_{k times (q-k)}$, $C_{(m-k) times k}$ and $D_{(m-k)times(q-k)}$, where the arrangement is done by performing $k$ row operations $E_1$ and $k$ column operations $E_2$ so that in $E_2E_1 mathbf{A}$, the minor $A$ is positioned at the upper left corner of the matrix, and $B, C, D$ are the other blocks. Since nonsingular matrices form an open subset, there is a neighborhood $U$ of $mathbf{A}$ in $M_{m times q}(Bbb R)$ where for any $mathbf{A}_0 in U$, the $(k times k)$-minor situated in the same position that $A$ is situated in $mathbf{A}$ is also nonsingular. Therefore, the blocks $A, B, C, D$ of $mathbf{A}$ are globally defined for every matrix in the open neighborhood $U$ of $mathbf{A}$.
Let $mathbf{X} in U$ be arbitrary and let $A, B, C, D$ be the blocks of $mathbf{X}$ as defined above. Notice that
$$mathbf{X} cdot begin{pmatrix} I & -A^{-1}B \ mathsf{O} & I end{pmatrix} = begin{pmatrix} A & mathsf{O} \ C & D - CA^{-1}B end{pmatrix}$$
Since the matrix we multiplied with is invertible, rank has remained unchanged. If $D - CA^{-1}B$ was nonzero we would have found a row in the bottom $(m-k)$-half of the matrix with a nonzero entry at one of the last $(q-k)$-th positions. This would then be linearly independent from the first $k$ rows of the matrix (as their last $(q-k)$ entries are all zero), which are themselves linearly independent by hypothesis on the minor $A$. That makes $text{rank}(mathbf{X}) geq k+1$, absurd. Therefore $D - CA^{-1}B = mathsf{O}$ is a necessary condition for $text{rank}(mathbf{X}) = k$. Once this condition holds the matrix above will have $k$ columns of vectors in $Bbb R^m$ whose projection to the first $q$ coordinates are linearly independent, hence are themselves linearly independent.
So consider the map $F : U to M_{(m-k)(q-k)}(Bbb R)$ given by $F(mathbf{X})= D - CA^{-1}B$. It's an easy check that $F$ is a smooth submersion, so $F^{-1}(mathsf{O}) = U cap Sigma_k$ is a submanifold of codimension $(m-k)(q-k)$ in $U$, which proves that $Sigma_k$ is a submanifold of $M_{m times q}(Bbb R)$.
This is not a closed, but a locally closed submanifold, of $M_{m times q}(Bbb R)$, and the closure $overline{Sigma_k}$ is precisely the subset of rank $leq k$ matrices in $M_{m times q}(Bbb R)$. This can be seen by taking any matrix $mathbf{M}$ with $text{rank}(mathbf{M}) = l < k$, which implies the $q$ column vectors $v_1, cdots, v_q$ span a subspace of dimension $l$ in $Bbb R^q$. Without loss of generality suppose $v_1, cdots, v_l$ is a basis of this subspace. Then $text{Span}(v_{l+1}), cdots, text{Span}
(v_{l+k})$ belong to this $l$-dimensional subspace; by a small perturbation in $Bbb R^q$ we can make them transverse to this $l$-dimensional subspace as well as with themselves. Let $w_{l+1}, cdots, w_{l+k}$ be the modified vectors; the matrix formed by the column vectors $v_1, cdots, v_l, w_{l+1}, cdots, w_{l+k}, v_{l+k+1}, cdots, v_q$ is a perturbation of $mathbf{M}$ which has rank $k$ (Note: if $k > q$, do the same argument with the $m$ rows). However, rank can never go down, as transverse subspaces remain transverse under small perturbations.
Therefore, $Sigma_k subset M_{m times q}(Bbb R)$ are locally closed submanifolds such that $overline{Sigma_k} = Sigma_0 cup cdots cup Sigma_{k-1}$. This kind of iterative structure is called a stratification; indeed, $Sigma_k$'s constitute stratums for a Whitney stratification of $M_{m times q}(Bbb R)$. But this was a long digression.
edited Dec 4 at 11:35
answered Dec 3 at 1:39
Balarka Sen
9,77912956
9,77912956
(+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
– loup blanc
Dec 4 at 11:33
add a comment |
(+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
– loup blanc
Dec 4 at 11:33
(+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
– loup blanc
Dec 4 at 11:33
(+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
– loup blanc
Dec 4 at 11:33
add a comment |
up vote
0
down vote
It is an open subspace of a vector space, $Gl(n,mathbb{R})=det^{-1}(mathbb{R}-{0}$. Here the vector space is the space of $ntimes n$-matrices so it is a submaniflod with one chart which is the embedding map $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.
add a comment |
up vote
0
down vote
It is an open subspace of a vector space, $Gl(n,mathbb{R})=det^{-1}(mathbb{R}-{0}$. Here the vector space is the space of $ntimes n$-matrices so it is a submaniflod with one chart which is the embedding map $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.
add a comment |
up vote
0
down vote
up vote
0
down vote
It is an open subspace of a vector space, $Gl(n,mathbb{R})=det^{-1}(mathbb{R}-{0}$. Here the vector space is the space of $ntimes n$-matrices so it is a submaniflod with one chart which is the embedding map $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.
It is an open subspace of a vector space, $Gl(n,mathbb{R})=det^{-1}(mathbb{R}-{0}$. Here the vector space is the space of $ntimes n$-matrices so it is a submaniflod with one chart which is the embedding map $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.
answered Dec 2 at 14:54
Tsemo Aristide
54.8k11444
54.8k11444
add a comment |
add a comment |
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