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.
$$det(M_n)=sum_{k=0}^mbinom{n}{2k}x^{n-2k}.$$







linear-algebra determinants







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.
$$det(M_n)=sum_{k=0}^mbinom{n}{2k}x^{n-2k}.$$







linear-algebra determinants







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.
$$det(M_n)=sum_{k=0}^mbinom{n}{2k}x^{n-2k}.$$







linear-algebra determinants







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




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

17.3k229127




17.3k229127












  • $begingroup$
    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













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1 Answer
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$begingroup$

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

9,08311851




9,08311851








  • 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




















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