Non singular matrices are a differentiable submanifold











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 $A$?










share|cite|improve this question


























    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 $A$?










    share|cite|improve this question
























      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 $A$?










      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 2 at 13:47









      hecho

      114




      114






















          2 Answers
          2






          active

          oldest

          votes

















          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.






          share|cite|improve this answer























          • (+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
            – loup blanc
            Dec 4 at 11:33


















          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 $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.






          share|cite|improve this answer





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022656%2fnon-singular-matrices-are-a-differentiable-submanifold%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            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.






            share|cite|improve this answer























            • (+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
              – loup blanc
              Dec 4 at 11:33















            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.






            share|cite|improve this answer























            • (+1) Clearly, a good answer; unfortunately, I'm afraid the OP did not understand much about it...
              – loup blanc
              Dec 4 at 11:33













            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.






            share|cite|improve this answer














            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            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


















            • (+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










            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 $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.






            share|cite|improve this answer

























              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 $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.






              share|cite|improve this answer























                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 $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.






                share|cite|improve this answer












                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 $Gl(n,mathbb{R})rightarrow M(n,mathbb{R})$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 at 14:54









                Tsemo Aristide

                54.8k11444




                54.8k11444






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022656%2fnon-singular-matrices-are-a-differentiable-submanifold%23new-answer', 'question_page');
                    }
                    );

                    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







                    Popular posts from this blog

                    Bressuire

                    Cabo Verde

                    Gyllenstierna