Circuits of matroid after adding rows to its representation
$begingroup$
I just started learning about matroids and oriented matroids, and my point of interest is their relation to the ideals of linear varieties/subspaces.
If I start with a proper linear subspace $L subset mathbb{R}^n$ spanned by column vectors $vec{u}$ and $vec{v}$, I can associate to it the matroid $M(L)$ represented by
$begin{bmatrix}vec{u}^T\ vec{v}^Tend{bmatrix}$
If I now take another vector $vec{w}$, and define a linear subspace $L + vec{w} = span{vec{u},vec{v},vec{w}}$, consider the matroid $M(L+vec{w})$: how does this matroid relate to $M(L)$, $L$, and $vec{w}$? For example, can its circuits be understood in terms of the circuits of $M(L)$, possibly as the union of certain circuits, determined by the particular relation of $vec{w}$ to $L$?
algebraic-geometry matroids
$endgroup$
add a comment |
$begingroup$
I just started learning about matroids and oriented matroids, and my point of interest is their relation to the ideals of linear varieties/subspaces.
If I start with a proper linear subspace $L subset mathbb{R}^n$ spanned by column vectors $vec{u}$ and $vec{v}$, I can associate to it the matroid $M(L)$ represented by
$begin{bmatrix}vec{u}^T\ vec{v}^Tend{bmatrix}$
If I now take another vector $vec{w}$, and define a linear subspace $L + vec{w} = span{vec{u},vec{v},vec{w}}$, consider the matroid $M(L+vec{w})$: how does this matroid relate to $M(L)$, $L$, and $vec{w}$? For example, can its circuits be understood in terms of the circuits of $M(L)$, possibly as the union of certain circuits, determined by the particular relation of $vec{w}$ to $L$?
algebraic-geometry matroids
$endgroup$
$begingroup$
This operation is typically called a one-element extension or a single-element extension. See Section 7.2 in Oxley's "Matroid Theory" and Section 7.1 in "Oriented Matroids" by Björner et al.
$endgroup$
– Aaron Dall
Feb 2 at 16:49
add a comment |
$begingroup$
I just started learning about matroids and oriented matroids, and my point of interest is their relation to the ideals of linear varieties/subspaces.
If I start with a proper linear subspace $L subset mathbb{R}^n$ spanned by column vectors $vec{u}$ and $vec{v}$, I can associate to it the matroid $M(L)$ represented by
$begin{bmatrix}vec{u}^T\ vec{v}^Tend{bmatrix}$
If I now take another vector $vec{w}$, and define a linear subspace $L + vec{w} = span{vec{u},vec{v},vec{w}}$, consider the matroid $M(L+vec{w})$: how does this matroid relate to $M(L)$, $L$, and $vec{w}$? For example, can its circuits be understood in terms of the circuits of $M(L)$, possibly as the union of certain circuits, determined by the particular relation of $vec{w}$ to $L$?
algebraic-geometry matroids
$endgroup$
I just started learning about matroids and oriented matroids, and my point of interest is their relation to the ideals of linear varieties/subspaces.
If I start with a proper linear subspace $L subset mathbb{R}^n$ spanned by column vectors $vec{u}$ and $vec{v}$, I can associate to it the matroid $M(L)$ represented by
$begin{bmatrix}vec{u}^T\ vec{v}^Tend{bmatrix}$
If I now take another vector $vec{w}$, and define a linear subspace $L + vec{w} = span{vec{u},vec{v},vec{w}}$, consider the matroid $M(L+vec{w})$: how does this matroid relate to $M(L)$, $L$, and $vec{w}$? For example, can its circuits be understood in terms of the circuits of $M(L)$, possibly as the union of certain circuits, determined by the particular relation of $vec{w}$ to $L$?
algebraic-geometry matroids
algebraic-geometry matroids
asked Jan 9 at 18:51
sw543sw543
283
283
$begingroup$
This operation is typically called a one-element extension or a single-element extension. See Section 7.2 in Oxley's "Matroid Theory" and Section 7.1 in "Oriented Matroids" by Björner et al.
$endgroup$
– Aaron Dall
Feb 2 at 16:49
add a comment |
$begingroup$
This operation is typically called a one-element extension or a single-element extension. See Section 7.2 in Oxley's "Matroid Theory" and Section 7.1 in "Oriented Matroids" by Björner et al.
$endgroup$
– Aaron Dall
Feb 2 at 16:49
$begingroup$
This operation is typically called a one-element extension or a single-element extension. See Section 7.2 in Oxley's "Matroid Theory" and Section 7.1 in "Oriented Matroids" by Björner et al.
$endgroup$
– Aaron Dall
Feb 2 at 16:49
$begingroup$
This operation is typically called a one-element extension or a single-element extension. See Section 7.2 in Oxley's "Matroid Theory" and Section 7.1 in "Oriented Matroids" by Björner et al.
$endgroup$
– Aaron Dall
Feb 2 at 16:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am assuming that the ground set of your matroids are $[n]$, corresponding to the columns of your matrix.
The notion relating $M(L+w)$ and $M(L)$ is a matroid quotient. Given matroids $M$ and $N$ on the same ground set, $M$ is a quotient of $N$, denoted $M to N$, if every flat of $N$ is also a flat of $M$. This is equivalent to requiring that every (union of) circuit(s) of $M$ is a union of circuits of $N$, as cocircuits are complements of hyperplanes and by duality.
Theorem: if $L subseteq L'$ are linear spaces, then $M(L') to M(L)$ is a matroid quotient.
Proof: Given a linear space $L subseteq mathbb{R}^n$, let $L^perp$ denote the orthogonal linear space with the standard dot produc. Also, let $supp: mathbb{R}^n to [n]$ be the map sending a vector to its support, i.e. the set of coordinates which are nonzero. Then the circuits of $M(L)$ are exactly the sets in $supp(L^perp) setminus varnothing$ which are minimal with respect to inclusion.1 In fact, every element of $supp(L^perp)$ is a union of circuits.2 Then if $L subseteq L'$, by taking orthogonal complements $L^perp supseteq L'^perp$, so $supp(L^perp) supseteq supp(L'^perp)$. Thus, every circuit of $M(L')$ is a union of circuits of $M(L)$, i.e. there is a matroid quotient $M(L') to M(L)$.
When a quotient $M to N$ has $rk, M - rk, N = 1$, these quotients are equivalent to so-called modular cuts. This story can be found in the "Constructions" chapter of Oxley's book.
The same story holds for oriented matroids if we replace the appropriate notions above with their signed analogues. In particular, support must be replaced by signed support, and union of circuits must be replaced by component-wise "addition" $circ$ defined by $+ circ x = +, 0 circ x = x, - circ - = -$. A similar story also holds for valuated matroids, and in general matroids over tracts.
1 Circuits correspond to minimal dependent sets, which for a vector matroid correspond to dependence relations (i.e. perpendicular vectors) of minimal support.
2 This can be proved by induction as follows: if $v in L^perp$ and $supp(v)$ is not minimal, pick $w in L^perp$ such that $supp(w) subseteq supp(v)$. Then for some $lambda$, we have $supp(v + lambda w)$ is strictly smaller than $supp(v)$ and can induct.
$endgroup$
$begingroup$
Thanks Joshua! I'll have to take some time to digest this, but this really helps.
$endgroup$
– sw543
Jan 10 at 14:46
add a comment |
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',
autoActivateHeartbeat: false,
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
});
}
});
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%2f3067821%2fcircuits-of-matroid-after-adding-rows-to-its-representation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I am assuming that the ground set of your matroids are $[n]$, corresponding to the columns of your matrix.
The notion relating $M(L+w)$ and $M(L)$ is a matroid quotient. Given matroids $M$ and $N$ on the same ground set, $M$ is a quotient of $N$, denoted $M to N$, if every flat of $N$ is also a flat of $M$. This is equivalent to requiring that every (union of) circuit(s) of $M$ is a union of circuits of $N$, as cocircuits are complements of hyperplanes and by duality.
Theorem: if $L subseteq L'$ are linear spaces, then $M(L') to M(L)$ is a matroid quotient.
Proof: Given a linear space $L subseteq mathbb{R}^n$, let $L^perp$ denote the orthogonal linear space with the standard dot produc. Also, let $supp: mathbb{R}^n to [n]$ be the map sending a vector to its support, i.e. the set of coordinates which are nonzero. Then the circuits of $M(L)$ are exactly the sets in $supp(L^perp) setminus varnothing$ which are minimal with respect to inclusion.1 In fact, every element of $supp(L^perp)$ is a union of circuits.2 Then if $L subseteq L'$, by taking orthogonal complements $L^perp supseteq L'^perp$, so $supp(L^perp) supseteq supp(L'^perp)$. Thus, every circuit of $M(L')$ is a union of circuits of $M(L)$, i.e. there is a matroid quotient $M(L') to M(L)$.
When a quotient $M to N$ has $rk, M - rk, N = 1$, these quotients are equivalent to so-called modular cuts. This story can be found in the "Constructions" chapter of Oxley's book.
The same story holds for oriented matroids if we replace the appropriate notions above with their signed analogues. In particular, support must be replaced by signed support, and union of circuits must be replaced by component-wise "addition" $circ$ defined by $+ circ x = +, 0 circ x = x, - circ - = -$. A similar story also holds for valuated matroids, and in general matroids over tracts.
1 Circuits correspond to minimal dependent sets, which for a vector matroid correspond to dependence relations (i.e. perpendicular vectors) of minimal support.
2 This can be proved by induction as follows: if $v in L^perp$ and $supp(v)$ is not minimal, pick $w in L^perp$ such that $supp(w) subseteq supp(v)$. Then for some $lambda$, we have $supp(v + lambda w)$ is strictly smaller than $supp(v)$ and can induct.
$endgroup$
$begingroup$
Thanks Joshua! I'll have to take some time to digest this, but this really helps.
$endgroup$
– sw543
Jan 10 at 14:46
add a comment |
$begingroup$
I am assuming that the ground set of your matroids are $[n]$, corresponding to the columns of your matrix.
The notion relating $M(L+w)$ and $M(L)$ is a matroid quotient. Given matroids $M$ and $N$ on the same ground set, $M$ is a quotient of $N$, denoted $M to N$, if every flat of $N$ is also a flat of $M$. This is equivalent to requiring that every (union of) circuit(s) of $M$ is a union of circuits of $N$, as cocircuits are complements of hyperplanes and by duality.
Theorem: if $L subseteq L'$ are linear spaces, then $M(L') to M(L)$ is a matroid quotient.
Proof: Given a linear space $L subseteq mathbb{R}^n$, let $L^perp$ denote the orthogonal linear space with the standard dot produc. Also, let $supp: mathbb{R}^n to [n]$ be the map sending a vector to its support, i.e. the set of coordinates which are nonzero. Then the circuits of $M(L)$ are exactly the sets in $supp(L^perp) setminus varnothing$ which are minimal with respect to inclusion.1 In fact, every element of $supp(L^perp)$ is a union of circuits.2 Then if $L subseteq L'$, by taking orthogonal complements $L^perp supseteq L'^perp$, so $supp(L^perp) supseteq supp(L'^perp)$. Thus, every circuit of $M(L')$ is a union of circuits of $M(L)$, i.e. there is a matroid quotient $M(L') to M(L)$.
When a quotient $M to N$ has $rk, M - rk, N = 1$, these quotients are equivalent to so-called modular cuts. This story can be found in the "Constructions" chapter of Oxley's book.
The same story holds for oriented matroids if we replace the appropriate notions above with their signed analogues. In particular, support must be replaced by signed support, and union of circuits must be replaced by component-wise "addition" $circ$ defined by $+ circ x = +, 0 circ x = x, - circ - = -$. A similar story also holds for valuated matroids, and in general matroids over tracts.
1 Circuits correspond to minimal dependent sets, which for a vector matroid correspond to dependence relations (i.e. perpendicular vectors) of minimal support.
2 This can be proved by induction as follows: if $v in L^perp$ and $supp(v)$ is not minimal, pick $w in L^perp$ such that $supp(w) subseteq supp(v)$. Then for some $lambda$, we have $supp(v + lambda w)$ is strictly smaller than $supp(v)$ and can induct.
$endgroup$
$begingroup$
Thanks Joshua! I'll have to take some time to digest this, but this really helps.
$endgroup$
– sw543
Jan 10 at 14:46
add a comment |
$begingroup$
I am assuming that the ground set of your matroids are $[n]$, corresponding to the columns of your matrix.
The notion relating $M(L+w)$ and $M(L)$ is a matroid quotient. Given matroids $M$ and $N$ on the same ground set, $M$ is a quotient of $N$, denoted $M to N$, if every flat of $N$ is also a flat of $M$. This is equivalent to requiring that every (union of) circuit(s) of $M$ is a union of circuits of $N$, as cocircuits are complements of hyperplanes and by duality.
Theorem: if $L subseteq L'$ are linear spaces, then $M(L') to M(L)$ is a matroid quotient.
Proof: Given a linear space $L subseteq mathbb{R}^n$, let $L^perp$ denote the orthogonal linear space with the standard dot produc. Also, let $supp: mathbb{R}^n to [n]$ be the map sending a vector to its support, i.e. the set of coordinates which are nonzero. Then the circuits of $M(L)$ are exactly the sets in $supp(L^perp) setminus varnothing$ which are minimal with respect to inclusion.1 In fact, every element of $supp(L^perp)$ is a union of circuits.2 Then if $L subseteq L'$, by taking orthogonal complements $L^perp supseteq L'^perp$, so $supp(L^perp) supseteq supp(L'^perp)$. Thus, every circuit of $M(L')$ is a union of circuits of $M(L)$, i.e. there is a matroid quotient $M(L') to M(L)$.
When a quotient $M to N$ has $rk, M - rk, N = 1$, these quotients are equivalent to so-called modular cuts. This story can be found in the "Constructions" chapter of Oxley's book.
The same story holds for oriented matroids if we replace the appropriate notions above with their signed analogues. In particular, support must be replaced by signed support, and union of circuits must be replaced by component-wise "addition" $circ$ defined by $+ circ x = +, 0 circ x = x, - circ - = -$. A similar story also holds for valuated matroids, and in general matroids over tracts.
1 Circuits correspond to minimal dependent sets, which for a vector matroid correspond to dependence relations (i.e. perpendicular vectors) of minimal support.
2 This can be proved by induction as follows: if $v in L^perp$ and $supp(v)$ is not minimal, pick $w in L^perp$ such that $supp(w) subseteq supp(v)$. Then for some $lambda$, we have $supp(v + lambda w)$ is strictly smaller than $supp(v)$ and can induct.
$endgroup$
I am assuming that the ground set of your matroids are $[n]$, corresponding to the columns of your matrix.
The notion relating $M(L+w)$ and $M(L)$ is a matroid quotient. Given matroids $M$ and $N$ on the same ground set, $M$ is a quotient of $N$, denoted $M to N$, if every flat of $N$ is also a flat of $M$. This is equivalent to requiring that every (union of) circuit(s) of $M$ is a union of circuits of $N$, as cocircuits are complements of hyperplanes and by duality.
Theorem: if $L subseteq L'$ are linear spaces, then $M(L') to M(L)$ is a matroid quotient.
Proof: Given a linear space $L subseteq mathbb{R}^n$, let $L^perp$ denote the orthogonal linear space with the standard dot produc. Also, let $supp: mathbb{R}^n to [n]$ be the map sending a vector to its support, i.e. the set of coordinates which are nonzero. Then the circuits of $M(L)$ are exactly the sets in $supp(L^perp) setminus varnothing$ which are minimal with respect to inclusion.1 In fact, every element of $supp(L^perp)$ is a union of circuits.2 Then if $L subseteq L'$, by taking orthogonal complements $L^perp supseteq L'^perp$, so $supp(L^perp) supseteq supp(L'^perp)$. Thus, every circuit of $M(L')$ is a union of circuits of $M(L)$, i.e. there is a matroid quotient $M(L') to M(L)$.
When a quotient $M to N$ has $rk, M - rk, N = 1$, these quotients are equivalent to so-called modular cuts. This story can be found in the "Constructions" chapter of Oxley's book.
The same story holds for oriented matroids if we replace the appropriate notions above with their signed analogues. In particular, support must be replaced by signed support, and union of circuits must be replaced by component-wise "addition" $circ$ defined by $+ circ x = +, 0 circ x = x, - circ - = -$. A similar story also holds for valuated matroids, and in general matroids over tracts.
1 Circuits correspond to minimal dependent sets, which for a vector matroid correspond to dependence relations (i.e. perpendicular vectors) of minimal support.
2 This can be proved by induction as follows: if $v in L^perp$ and $supp(v)$ is not minimal, pick $w in L^perp$ such that $supp(w) subseteq supp(v)$. Then for some $lambda$, we have $supp(v + lambda w)$ is strictly smaller than $supp(v)$ and can induct.
answered Jan 9 at 22:31
Joshua MundingerJoshua Mundinger
2,9171028
2,9171028
$begingroup$
Thanks Joshua! I'll have to take some time to digest this, but this really helps.
$endgroup$
– sw543
Jan 10 at 14:46
add a comment |
$begingroup$
Thanks Joshua! I'll have to take some time to digest this, but this really helps.
$endgroup$
– sw543
Jan 10 at 14:46
$begingroup$
Thanks Joshua! I'll have to take some time to digest this, but this really helps.
$endgroup$
– sw543
Jan 10 at 14:46
$begingroup$
Thanks Joshua! I'll have to take some time to digest this, but this really helps.
$endgroup$
– sw543
Jan 10 at 14:46
add a comment |
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.
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%2f3067821%2fcircuits-of-matroid-after-adding-rows-to-its-representation%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
$begingroup$
This operation is typically called a one-element extension or a single-element extension. See Section 7.2 in Oxley's "Matroid Theory" and Section 7.1 in "Oriented Matroids" by Björner et al.
$endgroup$
– Aaron Dall
Feb 2 at 16:49