Determinant of “skew-symmetric” matrices












7












$begingroup$


For $ninmathbb{N}$ and $m=lfloorfrac{n}2rfloor$, consider the $ntimes n$ skew-symmetric matrix $A_n$ where each entry in the first $m$
sub-diagonals below the main diagonal is $1$ and each of the remaining entries below the main diagonal is $-1$. Let $I_n$ be the $ntimes n$ identity matrix.



Next, construct the matrix $M_n=A_n+xI_n$. For example, we have
$$M_3=begin{pmatrix} x&-1&1 \ 1&x&-1 \ -1&1&x
end{pmatrix} qquad text{and} qquad
M_4=begin{pmatrix} x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
end{pmatrix}.$$




QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_{k=0}^mbinom{n}{2k}x^{n-2k}.$$











share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $A_n + x I_n$?
    $endgroup$
    – Gordon Royle
    Dec 12 '18 at 8:36










  • $begingroup$
    wouldn't $$det(M_n)=sum_{k=0}^m e_{n-2k}(x_1,dots,x_n)$$ be easier?
    $endgroup$
    – Martin Rubey
    Dec 12 '18 at 9:56






  • 1




    $begingroup$
    The coefficient of $x^{n-m}$ is the sum of determinants of $m times m$ principal submatrices of $A_n$. These are skew-symmetric, so have determinant zero when $m$ is odd. When $m=2k$ is even, such a determinant is a square of a Pfaffian, which (roughly speaking) counts weighted perfect matchings of, well, a size $2k$ subgraph of the oriented complete graph with oriented adjacency matrix $A_n$. I wonder if there is any combinatorial way to get $binom{n}{2k}$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
    $endgroup$
    – Zach Teitler
    Dec 12 '18 at 23:15
















7












$begingroup$


For $ninmathbb{N}$ and $m=lfloorfrac{n}2rfloor$, consider the $ntimes n$ skew-symmetric matrix $A_n$ where each entry in the first $m$
sub-diagonals below the main diagonal is $1$ and each of the remaining entries below the main diagonal is $-1$. Let $I_n$ be the $ntimes n$ identity matrix.



Next, construct the matrix $M_n=A_n+xI_n$. For example, we have
$$M_3=begin{pmatrix} x&-1&1 \ 1&x&-1 \ -1&1&x
end{pmatrix} qquad text{and} qquad
M_4=begin{pmatrix} x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
end{pmatrix}.$$




QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_{k=0}^mbinom{n}{2k}x^{n-2k}.$$











share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $A_n + x I_n$?
    $endgroup$
    – Gordon Royle
    Dec 12 '18 at 8:36










  • $begingroup$
    wouldn't $$det(M_n)=sum_{k=0}^m e_{n-2k}(x_1,dots,x_n)$$ be easier?
    $endgroup$
    – Martin Rubey
    Dec 12 '18 at 9:56






  • 1




    $begingroup$
    The coefficient of $x^{n-m}$ is the sum of determinants of $m times m$ principal submatrices of $A_n$. These are skew-symmetric, so have determinant zero when $m$ is odd. When $m=2k$ is even, such a determinant is a square of a Pfaffian, which (roughly speaking) counts weighted perfect matchings of, well, a size $2k$ subgraph of the oriented complete graph with oriented adjacency matrix $A_n$. I wonder if there is any combinatorial way to get $binom{n}{2k}$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
    $endgroup$
    – Zach Teitler
    Dec 12 '18 at 23:15














7












7








7


3



$begingroup$


For $ninmathbb{N}$ and $m=lfloorfrac{n}2rfloor$, consider the $ntimes n$ skew-symmetric matrix $A_n$ where each entry in the first $m$
sub-diagonals below the main diagonal is $1$ and each of the remaining entries below the main diagonal is $-1$. Let $I_n$ be the $ntimes n$ identity matrix.



Next, construct the matrix $M_n=A_n+xI_n$. For example, we have
$$M_3=begin{pmatrix} x&-1&1 \ 1&x&-1 \ -1&1&x
end{pmatrix} qquad text{and} qquad
M_4=begin{pmatrix} x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
end{pmatrix}.$$




QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_{k=0}^mbinom{n}{2k}x^{n-2k}.$$











share|cite|improve this question











$endgroup$




For $ninmathbb{N}$ and $m=lfloorfrac{n}2rfloor$, consider the $ntimes n$ skew-symmetric matrix $A_n$ where each entry in the first $m$
sub-diagonals below the main diagonal is $1$ and each of the remaining entries below the main diagonal is $-1$. Let $I_n$ be the $ntimes n$ identity matrix.



Next, construct the matrix $M_n=A_n+xI_n$. For example, we have
$$M_3=begin{pmatrix} x&-1&1 \ 1&x&-1 \ -1&1&x
end{pmatrix} qquad text{and} qquad
M_4=begin{pmatrix} x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
end{pmatrix}.$$




QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_{k=0}^mbinom{n}{2k}x^{n-2k}.$$








linear-algebra determinants






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 8:54







T. Amdeberhan

















asked Dec 12 '18 at 6:33









T. AmdeberhanT. Amdeberhan

17.5k229129




17.5k229129












  • $begingroup$
    Do you mean $A_n + x I_n$?
    $endgroup$
    – Gordon Royle
    Dec 12 '18 at 8:36










  • $begingroup$
    wouldn't $$det(M_n)=sum_{k=0}^m e_{n-2k}(x_1,dots,x_n)$$ be easier?
    $endgroup$
    – Martin Rubey
    Dec 12 '18 at 9:56






  • 1




    $begingroup$
    The coefficient of $x^{n-m}$ is the sum of determinants of $m times m$ principal submatrices of $A_n$. These are skew-symmetric, so have determinant zero when $m$ is odd. When $m=2k$ is even, such a determinant is a square of a Pfaffian, which (roughly speaking) counts weighted perfect matchings of, well, a size $2k$ subgraph of the oriented complete graph with oriented adjacency matrix $A_n$. I wonder if there is any combinatorial way to get $binom{n}{2k}$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
    $endgroup$
    – Zach Teitler
    Dec 12 '18 at 23:15


















  • $begingroup$
    Do you mean $A_n + x I_n$?
    $endgroup$
    – Gordon Royle
    Dec 12 '18 at 8:36










  • $begingroup$
    wouldn't $$det(M_n)=sum_{k=0}^m e_{n-2k}(x_1,dots,x_n)$$ be easier?
    $endgroup$
    – Martin Rubey
    Dec 12 '18 at 9:56






  • 1




    $begingroup$
    The coefficient of $x^{n-m}$ is the sum of determinants of $m times m$ principal submatrices of $A_n$. These are skew-symmetric, so have determinant zero when $m$ is odd. When $m=2k$ is even, such a determinant is a square of a Pfaffian, which (roughly speaking) counts weighted perfect matchings of, well, a size $2k$ subgraph of the oriented complete graph with oriented adjacency matrix $A_n$. I wonder if there is any combinatorial way to get $binom{n}{2k}$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
    $endgroup$
    – Zach Teitler
    Dec 12 '18 at 23:15
















$begingroup$
Do you mean $A_n + x I_n$?
$endgroup$
– Gordon Royle
Dec 12 '18 at 8:36




$begingroup$
Do you mean $A_n + x I_n$?
$endgroup$
– Gordon Royle
Dec 12 '18 at 8:36












$begingroup$
wouldn't $$det(M_n)=sum_{k=0}^m e_{n-2k}(x_1,dots,x_n)$$ be easier?
$endgroup$
– Martin Rubey
Dec 12 '18 at 9:56




$begingroup$
wouldn't $$det(M_n)=sum_{k=0}^m e_{n-2k}(x_1,dots,x_n)$$ be easier?
$endgroup$
– Martin Rubey
Dec 12 '18 at 9:56




1




1




$begingroup$
The coefficient of $x^{n-m}$ is the sum of determinants of $m times m$ principal submatrices of $A_n$. These are skew-symmetric, so have determinant zero when $m$ is odd. When $m=2k$ is even, such a determinant is a square of a Pfaffian, which (roughly speaking) counts weighted perfect matchings of, well, a size $2k$ subgraph of the oriented complete graph with oriented adjacency matrix $A_n$. I wonder if there is any combinatorial way to get $binom{n}{2k}$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
$endgroup$
– Zach Teitler
Dec 12 '18 at 23:15




$begingroup$
The coefficient of $x^{n-m}$ is the sum of determinants of $m times m$ principal submatrices of $A_n$. These are skew-symmetric, so have determinant zero when $m$ is odd. When $m=2k$ is even, such a determinant is a square of a Pfaffian, which (roughly speaking) counts weighted perfect matchings of, well, a size $2k$ subgraph of the oriented complete graph with oriented adjacency matrix $A_n$. I wonder if there is any combinatorial way to get $binom{n}{2k}$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
$endgroup$
– Zach Teitler
Dec 12 '18 at 23:15










1 Answer
1






active

oldest

votes


















2












$begingroup$

For $n$ odd, $M_n$ is an $ntimes n$ circulant matrix, and so Theorem 17 in Krattenthaler's marvellous text applies. Denoting by $w$ a primitive $n$th root of unity, it gives
$$det M_n=prod_{i=0}^{n-1} (x-w^i-w^{2i}-dots -w^{mi}+w^{(m+1)i}+dots +w^{(n-1)i}),$$
something that should not be hard to simplify.



For $n$ even, you still have a special Hankel matrix, for which again there are general methods in [loc.cit.].






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    For even $n$, there is an extension of circulant matrices sometimes called $alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$.
    $endgroup$
    – Federico Poloni
    Dec 12 '18 at 10:58













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: "504"
};
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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317466%2fdeterminant-of-skew-symmetric-matrices%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









2












$begingroup$

For $n$ odd, $M_n$ is an $ntimes n$ circulant matrix, and so Theorem 17 in Krattenthaler's marvellous text applies. Denoting by $w$ a primitive $n$th root of unity, it gives
$$det M_n=prod_{i=0}^{n-1} (x-w^i-w^{2i}-dots -w^{mi}+w^{(m+1)i}+dots +w^{(n-1)i}),$$
something that should not be hard to simplify.



For $n$ even, you still have a special Hankel matrix, for which again there are general methods in [loc.cit.].






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    For even $n$, there is an extension of circulant matrices sometimes called $alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$.
    $endgroup$
    – Federico Poloni
    Dec 12 '18 at 10:58


















2












$begingroup$

For $n$ odd, $M_n$ is an $ntimes n$ circulant matrix, and so Theorem 17 in Krattenthaler's marvellous text applies. Denoting by $w$ a primitive $n$th root of unity, it gives
$$det M_n=prod_{i=0}^{n-1} (x-w^i-w^{2i}-dots -w^{mi}+w^{(m+1)i}+dots +w^{(n-1)i}),$$
something that should not be hard to simplify.



For $n$ even, you still have a special Hankel matrix, for which again there are general methods in [loc.cit.].






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    For even $n$, there is an extension of circulant matrices sometimes called $alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$.
    $endgroup$
    – Federico Poloni
    Dec 12 '18 at 10:58
















2












2








2





$begingroup$

For $n$ odd, $M_n$ is an $ntimes n$ circulant matrix, and so Theorem 17 in Krattenthaler's marvellous text applies. Denoting by $w$ a primitive $n$th root of unity, it gives
$$det M_n=prod_{i=0}^{n-1} (x-w^i-w^{2i}-dots -w^{mi}+w^{(m+1)i}+dots +w^{(n-1)i}),$$
something that should not be hard to simplify.



For $n$ even, you still have a special Hankel matrix, for which again there are general methods in [loc.cit.].






share|cite|improve this answer









$endgroup$



For $n$ odd, $M_n$ is an $ntimes n$ circulant matrix, and so Theorem 17 in Krattenthaler's marvellous text applies. Denoting by $w$ a primitive $n$th root of unity, it gives
$$det M_n=prod_{i=0}^{n-1} (x-w^i-w^{2i}-dots -w^{mi}+w^{(m+1)i}+dots +w^{(n-1)i}),$$
something that should not be hard to simplify.



For $n$ even, you still have a special Hankel matrix, for which again there are general methods in [loc.cit.].







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 10:19









Dima PasechnikDima Pasechnik

9,10811851




9,10811851








  • 1




    $begingroup$
    For even $n$, there is an extension of circulant matrices sometimes called $alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$.
    $endgroup$
    – Federico Poloni
    Dec 12 '18 at 10:58
















  • 1




    $begingroup$
    For even $n$, there is an extension of circulant matrices sometimes called $alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$.
    $endgroup$
    – Federico Poloni
    Dec 12 '18 at 10:58










1




1




$begingroup$
For even $n$, there is an extension of circulant matrices sometimes called $alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$.
$endgroup$
– Federico Poloni
Dec 12 '18 at 10:58






$begingroup$
For even $n$, there is an extension of circulant matrices sometimes called $alpha$-circulant that should do the job; basically you can multiply the sub-diagonal entries by a certain complex number $alpha$. In this case I guess that the same formula holds, but with $w$ a root of $-1$ instead of $1$.
$endgroup$
– Federico Poloni
Dec 12 '18 at 10:58




















draft saved

draft discarded




















































Thanks for contributing an answer to MathOverflow!


  • 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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317466%2fdeterminant-of-skew-symmetric-matrices%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

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix