Determinantal symmetry: proof requested
$begingroup$
Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:
QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.
Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$
co.combinatorics linear-algebra determinants
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
$endgroup$
add a comment |
$begingroup$
Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:
QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.
Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$
co.combinatorics linear-algebra determinants
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
$endgroup$
4
$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago
2
$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago
1
$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago
add a comment |
$begingroup$
Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:
QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.
Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$
co.combinatorics linear-algebra determinants
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
$endgroup$
Consider the determinantal function
$$F(a,b,c):=detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}.$$
I would like to ask:
QUESTION. Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.
Example. The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$detbegin{pmatrix} 10&15&21&28\20&35&56&84\35&70&126&210\56&126&252&462
end{pmatrix}=
detbegin{pmatrix} 15&35&70\21&56&126\28&84&210
end{pmatrix}=detbegin{pmatrix} 35&56\70&126
end{pmatrix}.$$
co.combinatorics linear-algebra determinants
co.combinatorics linear-algebra determinants
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
edited 9 hours ago
T. Amdeberhan
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
asked 9 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
17.4k229127
4
$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago
2
$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago
1
$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago
add a comment |
4
$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago
2
$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago
1
$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago
4
4
$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago
$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago
2
2
$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago
$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago
1
1
$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago
$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
$endgroup$
add a comment |
$begingroup$
Just for the record:
In view of Fedor's nice reply, the interpretation gives away
$$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
frac{i+j+k+2}{i+j+k+1}.$$
An apparent symmetry!
As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.
Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
$$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
Then, we may generalize the above identity (1) as
$$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$
edited 6 hours ago
Fedor Petrov
49k5112229
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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$begingroup$
Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
$endgroup$
add a comment |
$begingroup$
Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
$endgroup$
add a comment |
$begingroup$
Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
$endgroup$
Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $atimes b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $atimes b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $atimes btimes c$ box. Thus the symmetry.
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
edited 6 hours ago
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
answered 9 hours ago
Fedor PetrovFedor Petrov
49k5112229
49k5112229
add a comment |
add a comment |
$begingroup$
Just for the record:
In view of Fedor's nice reply, the interpretation gives away
$$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
frac{i+j+k+2}{i+j+k+1}.$$
An apparent symmetry!
As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.
Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
$$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
Then, we may generalize the above identity (1) as
$$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$
edited 6 hours ago
Fedor Petrov
49k5112229
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
$endgroup$
add a comment |
$begingroup$
Just for the record:
In view of Fedor's nice reply, the interpretation gives away
$$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
frac{i+j+k+2}{i+j+k+1}.$$
An apparent symmetry!
As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.
Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
$$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
Then, we may generalize the above identity (1) as
$$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$
edited 6 hours ago
Fedor Petrov
49k5112229
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
$endgroup$
add a comment |
$begingroup$
Just for the record:
In view of Fedor's nice reply, the interpretation gives away
$$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
frac{i+j+k+2}{i+j+k+1}.$$
An apparent symmetry!
As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.
Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
$$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
Then, we may generalize the above identity (1) as
$$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$
edited 6 hours ago
Fedor Petrov
49k5112229
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
$endgroup$
Just for the record:
In view of Fedor's nice reply, the interpretation gives away
$$detleft[binom{i+j+a+b}{i+a}right]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}
frac{i+j+k+2}{i+j+k+1}.$$
An apparent symmetry!
As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.
Let $(q)_k=(1-q)(1-q^2)cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
$$binom{n}k_q=frac{(q)_n}{(q)_k(q)_{n-k}}.tag1$$
Then, we may generalize the above identity (1) as
$$detleft[q^{i^2+ai+bj}binom{i+j+a+b}{i+a}_qright]_{i,j=0}^{c-1}
=prod_{i=0}^{a-1}prod_{j=0}^{b-1}prod_{k=0}^{c-1}frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$
edited 6 hours ago
Fedor Petrov
49k5112229
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
edited 6 hours ago
Fedor Petrov
49k5112229
edited 6 hours ago
Fedor Petrov
49k5112229
edited 6 hours ago
Fedor Petrov
49k5112229
49k5112229
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
answered 8 hours ago
T. AmdeberhanT. Amdeberhan
17.4k229127
17.4k229127
add a comment |
add a comment |
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4
$begingroup$
Did you try to express this as the number of tuples of non-intersecting lattice paths?
$endgroup$
– Martin Rubey
9 hours ago
2
$begingroup$
The easy symmetry swapping $a$ and $b$ still holds if we replace the binomial coefficients with $q$-binomial coefficients. I tried some small cases with $a=2$, $b in {1,ldots, 10}$ and $c=3$, swapping $a$ and $c$, and, up to a power of $q$, there was still equality. In light of Fedor's answer, could this be explained by a $q$-analogue of the Lindstrom–Gessel–Viennot Lemma?
$endgroup$
– Mark Wildon
7 hours ago
1
$begingroup$
Or did you try anything low tech like column reduction and row reduction? Gerhard "Likes Easy To Understand Methods" Paseman, 2019.02.10.
$endgroup$
– Gerhard Paseman
7 hours ago