Determinant of “skew-symmetric” matrices
Clash Royale CLAN TAG#URR8PPP
For $ninmathbbN$ and $m=lfloorfracn2rfloor$, 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
$$M_3=beginpmatrix x&-1&1 \ 1&x&-1 \ -1&1&x
endpmatrix qquad textand qquad
M_4=beginpmatrix x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
endpmatrix.$$
QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_k=0^mbinomn2kx^n-2k.$$
linear-algebra determinants
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
add a comment |
For $ninmathbbN$ and $m=lfloorfracn2rfloor$, 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
$$M_3=beginpmatrix x&-1&1 \ 1&x&-1 \ -1&1&x
endpmatrix qquad textand qquad
M_4=beginpmatrix x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
endpmatrix.$$
QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_k=0^mbinomn2kx^n-2k.$$
linear-algebra determinants
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
Do you mean $A_n + x I_n$?
– Gordon Royle
Dec 12 at 8:36
wouldn't $$det(M_n)=sum_k=0^m e_n-2k(x_1,dots,x_n)$$ be easier?
– Martin Rubey
Dec 12 at 9:56
1
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 $binomn2k$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
– Zach Teitler
Dec 12 at 23:15
add a comment |
For $ninmathbbN$ and $m=lfloorfracn2rfloor$, 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
$$M_3=beginpmatrix x&-1&1 \ 1&x&-1 \ -1&1&x
endpmatrix qquad textand qquad
M_4=beginpmatrix x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
endpmatrix.$$
QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_k=0^mbinomn2kx^n-2k.$$
linear-algebra determinants
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
For $ninmathbbN$ and $m=lfloorfracn2rfloor$, 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
$$M_3=beginpmatrix x&-1&1 \ 1&x&-1 \ -1&1&x
endpmatrix qquad textand qquad
M_4=beginpmatrix x&-1&-1&1 \ 1&x&-1&-1 \ 1&1&x&-1 \ -1&1&1&x
endpmatrix.$$
QUESTION. Is the following true? Experiments suggest to be so.
$$det(M_n)=sum_k=0^mbinomn2kx^n-2k.$$
linear-algebra determinants
linear-algebra determinants
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
edited Dec 12 at 8:54
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
asked Dec 12 at 6:33
T. Amdeberhan
17k228126
17k228126
Do you mean $A_n + x I_n$?
– Gordon Royle
Dec 12 at 8:36
wouldn't $$det(M_n)=sum_k=0^m e_n-2k(x_1,dots,x_n)$$ be easier?
– Martin Rubey
Dec 12 at 9:56
1
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 $binomn2k$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
– Zach Teitler
Dec 12 at 23:15
add a comment |
Do you mean $A_n + x I_n$?
– Gordon Royle
Dec 12 at 8:36
wouldn't $$det(M_n)=sum_k=0^m e_n-2k(x_1,dots,x_n)$$ be easier?
– Martin Rubey
Dec 12 at 9:56
1
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 $binomn2k$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
– Zach Teitler
Dec 12 at 23:15
Do you mean $A_n + x I_n$?
– Gordon Royle
Dec 12 at 8:36
Do you mean $A_n + x I_n$?
– Gordon Royle
Dec 12 at 8:36
wouldn't $$det(M_n)=sum_k=0^m e_n-2k(x_1,dots,x_n)$$ be easier?
– Martin Rubey
Dec 12 at 9:56
wouldn't $$det(M_n)=sum_k=0^m e_n-2k(x_1,dots,x_n)$$ be easier?
– Martin Rubey
Dec 12 at 9:56
1
1
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 $binomn2k$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
– Zach Teitler
Dec 12 at 23:15
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 $binomn2k$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
– Zach Teitler
Dec 12 at 23:15
add a comment |
1 Answer
1
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oldest
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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.].
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
1
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 $1$.
– Federico Poloni
Dec 12 at 10:58
add a comment |
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oldest
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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.].
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
1
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 $1$.
– Federico Poloni
Dec 12 at 10:58
add a comment |
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.].
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
1
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 $1$.
– Federico Poloni
Dec 12 at 10:58
add a comment |
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.].
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
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.].
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
answered Dec 12 at 10:19
Dima Pasechnik
8,92111850
8,92111850
1
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 $1$.
– Federico Poloni
Dec 12 at 10:58
add a comment |
1
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 $1$.
– Federico Poloni
Dec 12 at 10:58
1
1
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 $1$.
– Federico Poloni
Dec 12 at 10:58
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 $1$.
– Federico Poloni
Dec 12 at 10:58
add a comment |
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Do you mean $A_n + x I_n$?
– Gordon Royle
Dec 12 at 8:36
wouldn't $$det(M_n)=sum_k=0^m e_n-2k(x_1,dots,x_n)$$ be easier?
– Martin Rubey
Dec 12 at 9:56
1
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 $binomn2k$ as the total of those squares of counts* of weighted matchings (*but they are not really counts, there are signs...).
– Zach Teitler
Dec 12 at 23:15