Mixing
Determinant is alternating over a commutative ring with $1$
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In Section 11.4 of Dummit and Foote, they introduce a determinant function $det$ on the ring of $ntimes n$ matrices over a commutative ring $R$ with $1$ as
Any $n$-multilinear alternating form, where the $n$-tuples are the $n$ columns of the matrices in $M_{ntimes n}(R)$, and
$det(I) = 1$ where $I$ is the $ntimes n$ identity matrix.
They then define a function
$$
det(alpha_{ij}) = sum_{sigmain S_n}operatorname{sgn}(sigma)alpha_{sigma(1)1}dotsbalpha_{sigma(n)n}
$$
and show that the determinant is unique, but they leave it as an exercise to show that the function $det$ defined above is actually a determinant function.
Note that Dummit and Foote take alternating to mean that if two consecutive columns of the matrix $(alpha_{ij})$ are equal, then the alternating form returns $0$ when applied to $(alpha_{ij})$.
I am having trouble showing that $det$ so-defined is alternating. I have managed to show that if a matrix $(alpha_{ij})$ has two consecutive columns equal, say the $j$th and $j+1$st, then $det(alpha_{ij})=-det(alpha_{ij})$. I am not sure if this is sufficient to show that $det(alpha_{ij}) = 0$ since we are in a commutative ring with $1$, which may have zero divisors.
Is there an easy fix? I can supply my proof if need be. Thanks.
linear-algebra abstract-algebra
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
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add a comment |
$begingroup$
In Section 11.4 of Dummit and Foote, they introduce a determinant function $det$ on the ring of $ntimes n$ matrices over a commutative ring $R$ with $1$ as
Any $n$-multilinear alternating form, where the $n$-tuples are the $n$ columns of the matrices in $M_{ntimes n}(R)$, and
$det(I) = 1$ where $I$ is the $ntimes n$ identity matrix.
They then define a function
$$
det(alpha_{ij}) = sum_{sigmain S_n}operatorname{sgn}(sigma)alpha_{sigma(1)1}dotsbalpha_{sigma(n)n}
$$
and show that the determinant is unique, but they leave it as an exercise to show that the function $det$ defined above is actually a determinant function.
Note that Dummit and Foote take alternating to mean that if two consecutive columns of the matrix $(alpha_{ij})$ are equal, then the alternating form returns $0$ when applied to $(alpha_{ij})$.
I am having trouble showing that $det$ so-defined is alternating. I have managed to show that if a matrix $(alpha_{ij})$ has two consecutive columns equal, say the $j$th and $j+1$st, then $det(alpha_{ij})=-det(alpha_{ij})$. I am not sure if this is sufficient to show that $det(alpha_{ij}) = 0$ since we are in a commutative ring with $1$, which may have zero divisors.
Is there an easy fix? I can supply my proof if need be. Thanks.
linear-algebra abstract-algebra
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
$endgroup$
add a comment |
$begingroup$
In Section 11.4 of Dummit and Foote, they introduce a determinant function $det$ on the ring of $ntimes n$ matrices over a commutative ring $R$ with $1$ as
Any $n$-multilinear alternating form, where the $n$-tuples are the $n$ columns of the matrices in $M_{ntimes n}(R)$, and
$det(I) = 1$ where $I$ is the $ntimes n$ identity matrix.
They then define a function
$$
det(alpha_{ij}) = sum_{sigmain S_n}operatorname{sgn}(sigma)alpha_{sigma(1)1}dotsbalpha_{sigma(n)n}
$$
and show that the determinant is unique, but they leave it as an exercise to show that the function $det$ defined above is actually a determinant function.
Note that Dummit and Foote take alternating to mean that if two consecutive columns of the matrix $(alpha_{ij})$ are equal, then the alternating form returns $0$ when applied to $(alpha_{ij})$.
I am having trouble showing that $det$ so-defined is alternating. I have managed to show that if a matrix $(alpha_{ij})$ has two consecutive columns equal, say the $j$th and $j+1$st, then $det(alpha_{ij})=-det(alpha_{ij})$. I am not sure if this is sufficient to show that $det(alpha_{ij}) = 0$ since we are in a commutative ring with $1$, which may have zero divisors.
Is there an easy fix? I can supply my proof if need be. Thanks.
linear-algebra abstract-algebra
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
$endgroup$
In Section 11.4 of Dummit and Foote, they introduce a determinant function $det$ on the ring of $ntimes n$ matrices over a commutative ring $R$ with $1$ as
Any $n$-multilinear alternating form, where the $n$-tuples are the $n$ columns of the matrices in $M_{ntimes n}(R)$, and
$det(I) = 1$ where $I$ is the $ntimes n$ identity matrix.
They then define a function
$$
det(alpha_{ij}) = sum_{sigmain S_n}operatorname{sgn}(sigma)alpha_{sigma(1)1}dotsbalpha_{sigma(n)n}
$$
and show that the determinant is unique, but they leave it as an exercise to show that the function $det$ defined above is actually a determinant function.
Note that Dummit and Foote take alternating to mean that if two consecutive columns of the matrix $(alpha_{ij})$ are equal, then the alternating form returns $0$ when applied to $(alpha_{ij})$.
I am having trouble showing that $det$ so-defined is alternating. I have managed to show that if a matrix $(alpha_{ij})$ has two consecutive columns equal, say the $j$th and $j+1$st, then $det(alpha_{ij})=-det(alpha_{ij})$. I am not sure if this is sufficient to show that $det(alpha_{ij}) = 0$ since we are in a commutative ring with $1$, which may have zero divisors.
Is there an easy fix? I can supply my proof if need be. Thanks.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
asked Jul 4 '17 at 16:39
Alex OrtizAlex Ortiz
11.5k21442
11.5k21442
add a comment |
add a comment |
3 Answers
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oldest
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You are correct. In the field with two elements (say), the equation $det(A) = -det(A)$ yields no new information--it might be best to argue the determinant is alternating directly.
Here's the idea for the 3x3 case, assuming that the first two columns of the matrix are equal. Let $G = S_3$ and let $H$ be the (two element) subgroup generated by the transposition $tau := (1, 2)$. Then partition $G$ into cosets $G/H$. In this specific example, $G/H = {{(1), tau}, {(1, 2, 3), (1, 2, 3)tau}, {(2,3), (2,3)tau}}$. Note that each coset has two permutations in it, some permutation, and then a transposition multiplied by that permutation.
Then recalling that the first two columns of the matrix are equal, you obtain that the determinant is $$sum_{sigma in S_3} sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H}sum_{sigma in S}sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma tau(1)1}a_{sigma tau(2)2}a_{sigma tau(3)3}) = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) stackrel{*}= sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)2}a_{sigma(2)1}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) = sum_{S in G/H, choose (sigma) in S} 0 = 0$$
where $stackrel{*}=$ uses the fact that the first two columns agree. (Note: the sum looks scary, but the idea is not nearly as scary.)
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
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Thanks for your answer, but I am not sure what you mean by suggesting to show that the determinant is alternating directly. In my mind, that is what I tried to show, but I came up with my problem.
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– Alex Ortiz
Jul 4 '17 at 17:41
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I mean you should show that after you switch two rows, each entry in that big sum is multiplied by -1 when computing the determinant of the new matrix.
$endgroup$
– Tom Gannon
Jul 4 '17 at 17:51
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Okay, but note that my definition of alternating is that when two adjacent columns (or rows) are equal, then the determinant function should evaluate to $0$. What you are suggesting doesn't seem to get me further in the right direction.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:54
$begingroup$
You're absolutely right, my apologies! I've updated my answer accordingly.
$endgroup$
– Tom Gannon
Jul 4 '17 at 18:29
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Ah, beautiful; I was able to turn this into a proof of the general case with no problem. I didn't think of partitioning $G$ into its cosets; does this idea come up in any other proofs you know of? Thanks very much!
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– Alex Ortiz
Jul 4 '17 at 19:01
|
show 3 more comments
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For a fully elaborated proof, I shall be lazy and just refer to Exercise 6.7 (e) in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. Or the proof of property (iii) in §5.3.4 of Hartmut Laue, Determinants. The main idea is to split the sum $sumlimits_{sigma in S_n} ldots$ into a sum over all even permutations and a sum over all odd permutations, and show that the addends in the two sums mutually cancel, as Tom Gannon suggests.
However, there is also a way to derive the result from your $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$ observation. Namely, fix $n in mathbb{N}$ and $k in left{1,2,ldots,n-1right}$. Let a $k$-col-equal matrix be a matrix whose $k$-th column equals its $k+1$-st column. Your claim is that a $k$-col-equal matrix must have determinant $0$. Now, observe that you can derive $u = 0$ from $u = -u$ when $u$ is a polynomial over $mathbb{Z}$ (for example). Thus, you can prove your claim whenever the entries of the $k$-col-equal matrix are polynomials over $mathbb{Z}$ (because you have shown that each $k$-col-equal matrix $left(alpha_{i,j}right)$ satisfies $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$). In particular, your claim holds when the col-equal matrix is the "universal $k$-col-equal matrix", which is the matrix whose entries are indeterminates (in a polynomial ring over $mathbb{Z}$) that are distinct except for the two columns that are supposed to be equal. (For example, the universal $2$-col-equal matrix for $n = 4$ is $left(begin{array}{cccc} x_{1,1} & x_{1,2} & x_{1,2} & x_{1,4} \ x_{2,1} & x_{2,2} & x_{2,2} & x_{2,4} \ x_{3,1} & x_{3,2} & x_{3,2} & x_{3,4} \ x_{4,1} & x_{4,2} & x_{4,2} & x_{4,4} end{array}right)$, where the $x_{i,j}$ are distinct indeterminates in a polynomial ring over $mathbb{Z}$.) But you can view an arbitrary $k$-col-equal matrix as a result of substituting concrete values for these indeterminates in the "universal $k$-col-equal matrix". Therefore, since your claim holds for the latter matrix, it must also hold for the former.
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
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This is a very interesting take on it. One that I had not considered, but that is very nice. It's good to know that the original result can still get us to the desired conclusion.
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– Alex Ortiz
Jul 23 '17 at 3:25
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Suppose we take a matrix $(alpha_{i,j})$, and switch the $p$th and $q$th rows to form another matrix $(beta_{i,j})$, which is to say,
$$beta_{i,j} = leftlbrace begin{array}{ccc} alpha_{i, p} &mathrm{if} & j = q \ alpha_{i, q} &mathrm{if} & j = p \ alpha_{i, j} & & mathrm{otherwise}end{array} right.$$
Another way to put it is, if we let $tau = (p ~ q) in S_n$, then,
$$beta_{i,j} = alpha_{i, tau(j)}$$
So we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{i in lbrace 1, ldots, n rbrace} alpha_{sigma(i),tau(i)}.$$
If we reindex the above product in terms of $j = tau(i) iff i = tau(j)$, then $j$ ranges over $lbrace 1, ldots, n rbrace$ as well (note the importance of commutativity in this step), and we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{j in lbrace 1, ldots, n rbrace} alpha_{sigma circ tau(j),j}.$$
We can now reindex the sum with $psi = sigma circ tau iff sigma = psi circ tau$, giving us
$$det(beta_{i,j}) = sum_{psi in S_n} operatorname{sgn}(psi circ tau) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j}.$$
But, since composing a permutation with a transposition changes its sign, we thus have,
$$det(beta_{i,j}) = -sum_{psi in S_n} operatorname{sgn}(psi) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j} = det(alpha_{i,j}).$$
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
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Do you mean to say switch the two columns? My indexing convention is that if I have $a_{ij}$, then $i$ denotes the row, and $j$ the column.
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– Alex Ortiz
Jul 4 '17 at 17:42
1
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Also, you have apparently shown what I have, which is that switching two rows (or columns) amounts to changing the sign. I am trying to show that if two neighboring columns (or rows) are identical, then the determinant is zero.
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– Alex Ortiz
Jul 4 '17 at 17:44
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3 Answers
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3 Answers
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You are correct. In the field with two elements (say), the equation $det(A) = -det(A)$ yields no new information--it might be best to argue the determinant is alternating directly.
Here's the idea for the 3x3 case, assuming that the first two columns of the matrix are equal. Let $G = S_3$ and let $H$ be the (two element) subgroup generated by the transposition $tau := (1, 2)$. Then partition $G$ into cosets $G/H$. In this specific example, $G/H = {{(1), tau}, {(1, 2, 3), (1, 2, 3)tau}, {(2,3), (2,3)tau}}$. Note that each coset has two permutations in it, some permutation, and then a transposition multiplied by that permutation.
Then recalling that the first two columns of the matrix are equal, you obtain that the determinant is $$sum_{sigma in S_3} sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H}sum_{sigma in S}sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma tau(1)1}a_{sigma tau(2)2}a_{sigma tau(3)3}) = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) stackrel{*}= sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)2}a_{sigma(2)1}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) = sum_{S in G/H, choose (sigma) in S} 0 = 0$$
where $stackrel{*}=$ uses the fact that the first two columns agree. (Note: the sum looks scary, but the idea is not nearly as scary.)
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
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$begingroup$
Thanks for your answer, but I am not sure what you mean by suggesting to show that the determinant is alternating directly. In my mind, that is what I tried to show, but I came up with my problem.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:41
$begingroup$
I mean you should show that after you switch two rows, each entry in that big sum is multiplied by -1 when computing the determinant of the new matrix.
$endgroup$
– Tom Gannon
Jul 4 '17 at 17:51
$begingroup$
Okay, but note that my definition of alternating is that when two adjacent columns (or rows) are equal, then the determinant function should evaluate to $0$. What you are suggesting doesn't seem to get me further in the right direction.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:54
$begingroup$
You're absolutely right, my apologies! I've updated my answer accordingly.
$endgroup$
– Tom Gannon
Jul 4 '17 at 18:29
$begingroup$
Ah, beautiful; I was able to turn this into a proof of the general case with no problem. I didn't think of partitioning $G$ into its cosets; does this idea come up in any other proofs you know of? Thanks very much!
$endgroup$
– Alex Ortiz
Jul 4 '17 at 19:01
|
show 3 more comments
$begingroup$
You are correct. In the field with two elements (say), the equation $det(A) = -det(A)$ yields no new information--it might be best to argue the determinant is alternating directly.
Here's the idea for the 3x3 case, assuming that the first two columns of the matrix are equal. Let $G = S_3$ and let $H$ be the (two element) subgroup generated by the transposition $tau := (1, 2)$. Then partition $G$ into cosets $G/H$. In this specific example, $G/H = {{(1), tau}, {(1, 2, 3), (1, 2, 3)tau}, {(2,3), (2,3)tau}}$. Note that each coset has two permutations in it, some permutation, and then a transposition multiplied by that permutation.
Then recalling that the first two columns of the matrix are equal, you obtain that the determinant is $$sum_{sigma in S_3} sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H}sum_{sigma in S}sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma tau(1)1}a_{sigma tau(2)2}a_{sigma tau(3)3}) = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) stackrel{*}= sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)2}a_{sigma(2)1}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) = sum_{S in G/H, choose (sigma) in S} 0 = 0$$
where $stackrel{*}=$ uses the fact that the first two columns agree. (Note: the sum looks scary, but the idea is not nearly as scary.)
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
$endgroup$
$begingroup$
Thanks for your answer, but I am not sure what you mean by suggesting to show that the determinant is alternating directly. In my mind, that is what I tried to show, but I came up with my problem.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:41
$begingroup$
I mean you should show that after you switch two rows, each entry in that big sum is multiplied by -1 when computing the determinant of the new matrix.
$endgroup$
– Tom Gannon
Jul 4 '17 at 17:51
$begingroup$
Okay, but note that my definition of alternating is that when two adjacent columns (or rows) are equal, then the determinant function should evaluate to $0$. What you are suggesting doesn't seem to get me further in the right direction.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:54
$begingroup$
You're absolutely right, my apologies! I've updated my answer accordingly.
$endgroup$
– Tom Gannon
Jul 4 '17 at 18:29
$begingroup$
Ah, beautiful; I was able to turn this into a proof of the general case with no problem. I didn't think of partitioning $G$ into its cosets; does this idea come up in any other proofs you know of? Thanks very much!
$endgroup$
– Alex Ortiz
Jul 4 '17 at 19:01
|
show 3 more comments
$begingroup$
You are correct. In the field with two elements (say), the equation $det(A) = -det(A)$ yields no new information--it might be best to argue the determinant is alternating directly.
Here's the idea for the 3x3 case, assuming that the first two columns of the matrix are equal. Let $G = S_3$ and let $H$ be the (two element) subgroup generated by the transposition $tau := (1, 2)$. Then partition $G$ into cosets $G/H$. In this specific example, $G/H = {{(1), tau}, {(1, 2, 3), (1, 2, 3)tau}, {(2,3), (2,3)tau}}$. Note that each coset has two permutations in it, some permutation, and then a transposition multiplied by that permutation.
Then recalling that the first two columns of the matrix are equal, you obtain that the determinant is $$sum_{sigma in S_3} sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H}sum_{sigma in S}sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma tau(1)1}a_{sigma tau(2)2}a_{sigma tau(3)3}) = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) stackrel{*}= sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)2}a_{sigma(2)1}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) = sum_{S in G/H, choose (sigma) in S} 0 = 0$$
where $stackrel{*}=$ uses the fact that the first two columns agree. (Note: the sum looks scary, but the idea is not nearly as scary.)
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
$endgroup$
You are correct. In the field with two elements (say), the equation $det(A) = -det(A)$ yields no new information--it might be best to argue the determinant is alternating directly.
Here's the idea for the 3x3 case, assuming that the first two columns of the matrix are equal. Let $G = S_3$ and let $H$ be the (two element) subgroup generated by the transposition $tau := (1, 2)$. Then partition $G$ into cosets $G/H$. In this specific example, $G/H = {{(1), tau}, {(1, 2, 3), (1, 2, 3)tau}, {(2,3), (2,3)tau}}$. Note that each coset has two permutations in it, some permutation, and then a transposition multiplied by that permutation.
Then recalling that the first two columns of the matrix are equal, you obtain that the determinant is $$sum_{sigma in S_3} sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H}sum_{sigma in S}sign(sigma)a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma tau(1)1}a_{sigma tau(2)2}a_{sigma tau(3)3}) = sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)1}a_{sigma(2)2}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) stackrel{*}= sum_{S in G/H, choose (sigma) in S} (pm a_{sigma(1)2}a_{sigma(2)1}a_{sigma(3)3} mp a_{sigma (2)1}a_{sigma (1)2}a_{sigma (3)3}) = sum_{S in G/H, choose (sigma) in S} 0 = 0$$
where $stackrel{*}=$ uses the fact that the first two columns agree. (Note: the sum looks scary, but the idea is not nearly as scary.)
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
edited Jul 22 '17 at 21:26
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
answered Jul 4 '17 at 17:34
Tom GannonTom Gannon
707210
707210
$begingroup$
Thanks for your answer, but I am not sure what you mean by suggesting to show that the determinant is alternating directly. In my mind, that is what I tried to show, but I came up with my problem.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:41
$begingroup$
I mean you should show that after you switch two rows, each entry in that big sum is multiplied by -1 when computing the determinant of the new matrix.
$endgroup$
– Tom Gannon
Jul 4 '17 at 17:51
$begingroup$
Okay, but note that my definition of alternating is that when two adjacent columns (or rows) are equal, then the determinant function should evaluate to $0$. What you are suggesting doesn't seem to get me further in the right direction.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:54
$begingroup$
You're absolutely right, my apologies! I've updated my answer accordingly.
$endgroup$
– Tom Gannon
Jul 4 '17 at 18:29
$begingroup$
Ah, beautiful; I was able to turn this into a proof of the general case with no problem. I didn't think of partitioning $G$ into its cosets; does this idea come up in any other proofs you know of? Thanks very much!
$endgroup$
– Alex Ortiz
Jul 4 '17 at 19:01
|
show 3 more comments
$begingroup$
Thanks for your answer, but I am not sure what you mean by suggesting to show that the determinant is alternating directly. In my mind, that is what I tried to show, but I came up with my problem.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:41
$begingroup$
I mean you should show that after you switch two rows, each entry in that big sum is multiplied by -1 when computing the determinant of the new matrix.
$endgroup$
– Tom Gannon
Jul 4 '17 at 17:51
$begingroup$
Okay, but note that my definition of alternating is that when two adjacent columns (or rows) are equal, then the determinant function should evaluate to $0$. What you are suggesting doesn't seem to get me further in the right direction.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:54
$begingroup$
You're absolutely right, my apologies! I've updated my answer accordingly.
$endgroup$
– Tom Gannon
Jul 4 '17 at 18:29
$begingroup$
Ah, beautiful; I was able to turn this into a proof of the general case with no problem. I didn't think of partitioning $G$ into its cosets; does this idea come up in any other proofs you know of? Thanks very much!
$endgroup$
– Alex Ortiz
Jul 4 '17 at 19:01
$begingroup$
Thanks for your answer, but I am not sure what you mean by suggesting to show that the determinant is alternating directly. In my mind, that is what I tried to show, but I came up with my problem.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:41
$begingroup$
Thanks for your answer, but I am not sure what you mean by suggesting to show that the determinant is alternating directly. In my mind, that is what I tried to show, but I came up with my problem.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:41
$begingroup$
I mean you should show that after you switch two rows, each entry in that big sum is multiplied by -1 when computing the determinant of the new matrix.
$endgroup$
– Tom Gannon
Jul 4 '17 at 17:51
$begingroup$
I mean you should show that after you switch two rows, each entry in that big sum is multiplied by -1 when computing the determinant of the new matrix.
$endgroup$
– Tom Gannon
Jul 4 '17 at 17:51
$begingroup$
Okay, but note that my definition of alternating is that when two adjacent columns (or rows) are equal, then the determinant function should evaluate to $0$. What you are suggesting doesn't seem to get me further in the right direction.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:54
$begingroup$
Okay, but note that my definition of alternating is that when two adjacent columns (or rows) are equal, then the determinant function should evaluate to $0$. What you are suggesting doesn't seem to get me further in the right direction.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:54
$begingroup$
You're absolutely right, my apologies! I've updated my answer accordingly.
$endgroup$
– Tom Gannon
Jul 4 '17 at 18:29
$begingroup$
You're absolutely right, my apologies! I've updated my answer accordingly.
$endgroup$
– Tom Gannon
Jul 4 '17 at 18:29
$begingroup$
Ah, beautiful; I was able to turn this into a proof of the general case with no problem. I didn't think of partitioning $G$ into its cosets; does this idea come up in any other proofs you know of? Thanks very much!
$endgroup$
– Alex Ortiz
Jul 4 '17 at 19:01
$begingroup$
Ah, beautiful; I was able to turn this into a proof of the general case with no problem. I didn't think of partitioning $G$ into its cosets; does this idea come up in any other proofs you know of? Thanks very much!
$endgroup$
– Alex Ortiz
Jul 4 '17 at 19:01
|
show 3 more comments
$begingroup$
For a fully elaborated proof, I shall be lazy and just refer to Exercise 6.7 (e) in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. Or the proof of property (iii) in §5.3.4 of Hartmut Laue, Determinants. The main idea is to split the sum $sumlimits_{sigma in S_n} ldots$ into a sum over all even permutations and a sum over all odd permutations, and show that the addends in the two sums mutually cancel, as Tom Gannon suggests.
However, there is also a way to derive the result from your $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$ observation. Namely, fix $n in mathbb{N}$ and $k in left{1,2,ldots,n-1right}$. Let a $k$-col-equal matrix be a matrix whose $k$-th column equals its $k+1$-st column. Your claim is that a $k$-col-equal matrix must have determinant $0$. Now, observe that you can derive $u = 0$ from $u = -u$ when $u$ is a polynomial over $mathbb{Z}$ (for example). Thus, you can prove your claim whenever the entries of the $k$-col-equal matrix are polynomials over $mathbb{Z}$ (because you have shown that each $k$-col-equal matrix $left(alpha_{i,j}right)$ satisfies $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$). In particular, your claim holds when the col-equal matrix is the "universal $k$-col-equal matrix", which is the matrix whose entries are indeterminates (in a polynomial ring over $mathbb{Z}$) that are distinct except for the two columns that are supposed to be equal. (For example, the universal $2$-col-equal matrix for $n = 4$ is $left(begin{array}{cccc} x_{1,1} & x_{1,2} & x_{1,2} & x_{1,4} \ x_{2,1} & x_{2,2} & x_{2,2} & x_{2,4} \ x_{3,1} & x_{3,2} & x_{3,2} & x_{3,4} \ x_{4,1} & x_{4,2} & x_{4,2} & x_{4,4} end{array}right)$, where the $x_{i,j}$ are distinct indeterminates in a polynomial ring over $mathbb{Z}$.) But you can view an arbitrary $k$-col-equal matrix as a result of substituting concrete values for these indeterminates in the "universal $k$-col-equal matrix". Therefore, since your claim holds for the latter matrix, it must also hold for the former.
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
$endgroup$
$begingroup$
This is a very interesting take on it. One that I had not considered, but that is very nice. It's good to know that the original result can still get us to the desired conclusion.
$endgroup$
– Alex Ortiz
Jul 23 '17 at 3:25
add a comment |
$begingroup$
For a fully elaborated proof, I shall be lazy and just refer to Exercise 6.7 (e) in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. Or the proof of property (iii) in §5.3.4 of Hartmut Laue, Determinants. The main idea is to split the sum $sumlimits_{sigma in S_n} ldots$ into a sum over all even permutations and a sum over all odd permutations, and show that the addends in the two sums mutually cancel, as Tom Gannon suggests.
However, there is also a way to derive the result from your $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$ observation. Namely, fix $n in mathbb{N}$ and $k in left{1,2,ldots,n-1right}$. Let a $k$-col-equal matrix be a matrix whose $k$-th column equals its $k+1$-st column. Your claim is that a $k$-col-equal matrix must have determinant $0$. Now, observe that you can derive $u = 0$ from $u = -u$ when $u$ is a polynomial over $mathbb{Z}$ (for example). Thus, you can prove your claim whenever the entries of the $k$-col-equal matrix are polynomials over $mathbb{Z}$ (because you have shown that each $k$-col-equal matrix $left(alpha_{i,j}right)$ satisfies $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$). In particular, your claim holds when the col-equal matrix is the "universal $k$-col-equal matrix", which is the matrix whose entries are indeterminates (in a polynomial ring over $mathbb{Z}$) that are distinct except for the two columns that are supposed to be equal. (For example, the universal $2$-col-equal matrix for $n = 4$ is $left(begin{array}{cccc} x_{1,1} & x_{1,2} & x_{1,2} & x_{1,4} \ x_{2,1} & x_{2,2} & x_{2,2} & x_{2,4} \ x_{3,1} & x_{3,2} & x_{3,2} & x_{3,4} \ x_{4,1} & x_{4,2} & x_{4,2} & x_{4,4} end{array}right)$, where the $x_{i,j}$ are distinct indeterminates in a polynomial ring over $mathbb{Z}$.) But you can view an arbitrary $k$-col-equal matrix as a result of substituting concrete values for these indeterminates in the "universal $k$-col-equal matrix". Therefore, since your claim holds for the latter matrix, it must also hold for the former.
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
$endgroup$
$begingroup$
This is a very interesting take on it. One that I had not considered, but that is very nice. It's good to know that the original result can still get us to the desired conclusion.
$endgroup$
– Alex Ortiz
Jul 23 '17 at 3:25
add a comment |
$begingroup$
For a fully elaborated proof, I shall be lazy and just refer to Exercise 6.7 (e) in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. Or the proof of property (iii) in §5.3.4 of Hartmut Laue, Determinants. The main idea is to split the sum $sumlimits_{sigma in S_n} ldots$ into a sum over all even permutations and a sum over all odd permutations, and show that the addends in the two sums mutually cancel, as Tom Gannon suggests.
However, there is also a way to derive the result from your $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$ observation. Namely, fix $n in mathbb{N}$ and $k in left{1,2,ldots,n-1right}$. Let a $k$-col-equal matrix be a matrix whose $k$-th column equals its $k+1$-st column. Your claim is that a $k$-col-equal matrix must have determinant $0$. Now, observe that you can derive $u = 0$ from $u = -u$ when $u$ is a polynomial over $mathbb{Z}$ (for example). Thus, you can prove your claim whenever the entries of the $k$-col-equal matrix are polynomials over $mathbb{Z}$ (because you have shown that each $k$-col-equal matrix $left(alpha_{i,j}right)$ satisfies $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$). In particular, your claim holds when the col-equal matrix is the "universal $k$-col-equal matrix", which is the matrix whose entries are indeterminates (in a polynomial ring over $mathbb{Z}$) that are distinct except for the two columns that are supposed to be equal. (For example, the universal $2$-col-equal matrix for $n = 4$ is $left(begin{array}{cccc} x_{1,1} & x_{1,2} & x_{1,2} & x_{1,4} \ x_{2,1} & x_{2,2} & x_{2,2} & x_{2,4} \ x_{3,1} & x_{3,2} & x_{3,2} & x_{3,4} \ x_{4,1} & x_{4,2} & x_{4,2} & x_{4,4} end{array}right)$, where the $x_{i,j}$ are distinct indeterminates in a polynomial ring over $mathbb{Z}$.) But you can view an arbitrary $k$-col-equal matrix as a result of substituting concrete values for these indeterminates in the "universal $k$-col-equal matrix". Therefore, since your claim holds for the latter matrix, it must also hold for the former.
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
$endgroup$
For a fully elaborated proof, I shall be lazy and just refer to Exercise 6.7 (e) in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. Or the proof of property (iii) in §5.3.4 of Hartmut Laue, Determinants. The main idea is to split the sum $sumlimits_{sigma in S_n} ldots$ into a sum over all even permutations and a sum over all odd permutations, and show that the addends in the two sums mutually cancel, as Tom Gannon suggests.
However, there is also a way to derive the result from your $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$ observation. Namely, fix $n in mathbb{N}$ and $k in left{1,2,ldots,n-1right}$. Let a $k$-col-equal matrix be a matrix whose $k$-th column equals its $k+1$-st column. Your claim is that a $k$-col-equal matrix must have determinant $0$. Now, observe that you can derive $u = 0$ from $u = -u$ when $u$ is a polynomial over $mathbb{Z}$ (for example). Thus, you can prove your claim whenever the entries of the $k$-col-equal matrix are polynomials over $mathbb{Z}$ (because you have shown that each $k$-col-equal matrix $left(alpha_{i,j}right)$ satisfies $det left(alpha_{i,j}right) = - det left(alpha_{i,j}right)$). In particular, your claim holds when the col-equal matrix is the "universal $k$-col-equal matrix", which is the matrix whose entries are indeterminates (in a polynomial ring over $mathbb{Z}$) that are distinct except for the two columns that are supposed to be equal. (For example, the universal $2$-col-equal matrix for $n = 4$ is $left(begin{array}{cccc} x_{1,1} & x_{1,2} & x_{1,2} & x_{1,4} \ x_{2,1} & x_{2,2} & x_{2,2} & x_{2,4} \ x_{3,1} & x_{3,2} & x_{3,2} & x_{3,4} \ x_{4,1} & x_{4,2} & x_{4,2} & x_{4,4} end{array}right)$, where the $x_{i,j}$ are distinct indeterminates in a polynomial ring over $mathbb{Z}$.) But you can view an arbitrary $k$-col-equal matrix as a result of substituting concrete values for these indeterminates in the "universal $k$-col-equal matrix". Therefore, since your claim holds for the latter matrix, it must also hold for the former.
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
edited Feb 3 at 6:29
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
answered Jul 22 '17 at 22:36
darij grinbergdarij grinberg
11.5k33168
11.5k33168
$begingroup$
This is a very interesting take on it. One that I had not considered, but that is very nice. It's good to know that the original result can still get us to the desired conclusion.
$endgroup$
– Alex Ortiz
Jul 23 '17 at 3:25
add a comment |
$begingroup$
This is a very interesting take on it. One that I had not considered, but that is very nice. It's good to know that the original result can still get us to the desired conclusion.
$endgroup$
– Alex Ortiz
Jul 23 '17 at 3:25
$begingroup$
This is a very interesting take on it. One that I had not considered, but that is very nice. It's good to know that the original result can still get us to the desired conclusion.
$endgroup$
– Alex Ortiz
Jul 23 '17 at 3:25
$begingroup$
This is a very interesting take on it. One that I had not considered, but that is very nice. It's good to know that the original result can still get us to the desired conclusion.
$endgroup$
– Alex Ortiz
Jul 23 '17 at 3:25
add a comment |
$begingroup$
Suppose we take a matrix $(alpha_{i,j})$, and switch the $p$th and $q$th rows to form another matrix $(beta_{i,j})$, which is to say,
$$beta_{i,j} = leftlbrace begin{array}{ccc} alpha_{i, p} &mathrm{if} & j = q \ alpha_{i, q} &mathrm{if} & j = p \ alpha_{i, j} & & mathrm{otherwise}end{array} right.$$
Another way to put it is, if we let $tau = (p ~ q) in S_n$, then,
$$beta_{i,j} = alpha_{i, tau(j)}$$
So we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{i in lbrace 1, ldots, n rbrace} alpha_{sigma(i),tau(i)}.$$
If we reindex the above product in terms of $j = tau(i) iff i = tau(j)$, then $j$ ranges over $lbrace 1, ldots, n rbrace$ as well (note the importance of commutativity in this step), and we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{j in lbrace 1, ldots, n rbrace} alpha_{sigma circ tau(j),j}.$$
We can now reindex the sum with $psi = sigma circ tau iff sigma = psi circ tau$, giving us
$$det(beta_{i,j}) = sum_{psi in S_n} operatorname{sgn}(psi circ tau) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j}.$$
But, since composing a permutation with a transposition changes its sign, we thus have,
$$det(beta_{i,j}) = -sum_{psi in S_n} operatorname{sgn}(psi) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j} = det(alpha_{i,j}).$$
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
$endgroup$
$begingroup$
Do you mean to say switch the two columns? My indexing convention is that if I have $a_{ij}$, then $i$ denotes the row, and $j$ the column.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:42
1
$begingroup$
Also, you have apparently shown what I have, which is that switching two rows (or columns) amounts to changing the sign. I am trying to show that if two neighboring columns (or rows) are identical, then the determinant is zero.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:44
add a comment |
$begingroup$
Suppose we take a matrix $(alpha_{i,j})$, and switch the $p$th and $q$th rows to form another matrix $(beta_{i,j})$, which is to say,
$$beta_{i,j} = leftlbrace begin{array}{ccc} alpha_{i, p} &mathrm{if} & j = q \ alpha_{i, q} &mathrm{if} & j = p \ alpha_{i, j} & & mathrm{otherwise}end{array} right.$$
Another way to put it is, if we let $tau = (p ~ q) in S_n$, then,
$$beta_{i,j} = alpha_{i, tau(j)}$$
So we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{i in lbrace 1, ldots, n rbrace} alpha_{sigma(i),tau(i)}.$$
If we reindex the above product in terms of $j = tau(i) iff i = tau(j)$, then $j$ ranges over $lbrace 1, ldots, n rbrace$ as well (note the importance of commutativity in this step), and we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{j in lbrace 1, ldots, n rbrace} alpha_{sigma circ tau(j),j}.$$
We can now reindex the sum with $psi = sigma circ tau iff sigma = psi circ tau$, giving us
$$det(beta_{i,j}) = sum_{psi in S_n} operatorname{sgn}(psi circ tau) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j}.$$
But, since composing a permutation with a transposition changes its sign, we thus have,
$$det(beta_{i,j}) = -sum_{psi in S_n} operatorname{sgn}(psi) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j} = det(alpha_{i,j}).$$
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
$endgroup$
$begingroup$
Do you mean to say switch the two columns? My indexing convention is that if I have $a_{ij}$, then $i$ denotes the row, and $j$ the column.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:42
1
$begingroup$
Also, you have apparently shown what I have, which is that switching two rows (or columns) amounts to changing the sign. I am trying to show that if two neighboring columns (or rows) are identical, then the determinant is zero.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:44
add a comment |
$begingroup$
Suppose we take a matrix $(alpha_{i,j})$, and switch the $p$th and $q$th rows to form another matrix $(beta_{i,j})$, which is to say,
$$beta_{i,j} = leftlbrace begin{array}{ccc} alpha_{i, p} &mathrm{if} & j = q \ alpha_{i, q} &mathrm{if} & j = p \ alpha_{i, j} & & mathrm{otherwise}end{array} right.$$
Another way to put it is, if we let $tau = (p ~ q) in S_n$, then,
$$beta_{i,j} = alpha_{i, tau(j)}$$
So we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{i in lbrace 1, ldots, n rbrace} alpha_{sigma(i),tau(i)}.$$
If we reindex the above product in terms of $j = tau(i) iff i = tau(j)$, then $j$ ranges over $lbrace 1, ldots, n rbrace$ as well (note the importance of commutativity in this step), and we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{j in lbrace 1, ldots, n rbrace} alpha_{sigma circ tau(j),j}.$$
We can now reindex the sum with $psi = sigma circ tau iff sigma = psi circ tau$, giving us
$$det(beta_{i,j}) = sum_{psi in S_n} operatorname{sgn}(psi circ tau) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j}.$$
But, since composing a permutation with a transposition changes its sign, we thus have,
$$det(beta_{i,j}) = -sum_{psi in S_n} operatorname{sgn}(psi) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j} = det(alpha_{i,j}).$$
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
$endgroup$
Suppose we take a matrix $(alpha_{i,j})$, and switch the $p$th and $q$th rows to form another matrix $(beta_{i,j})$, which is to say,
$$beta_{i,j} = leftlbrace begin{array}{ccc} alpha_{i, p} &mathrm{if} & j = q \ alpha_{i, q} &mathrm{if} & j = p \ alpha_{i, j} & & mathrm{otherwise}end{array} right.$$
Another way to put it is, if we let $tau = (p ~ q) in S_n$, then,
$$beta_{i,j} = alpha_{i, tau(j)}$$
So we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{i in lbrace 1, ldots, n rbrace} alpha_{sigma(i),tau(i)}.$$
If we reindex the above product in terms of $j = tau(i) iff i = tau(j)$, then $j$ ranges over $lbrace 1, ldots, n rbrace$ as well (note the importance of commutativity in this step), and we have,
$$det(beta_{i,j}) = sum_{sigma in S_n} operatorname{sgn}(sigma) prod_{j in lbrace 1, ldots, n rbrace} alpha_{sigma circ tau(j),j}.$$
We can now reindex the sum with $psi = sigma circ tau iff sigma = psi circ tau$, giving us
$$det(beta_{i,j}) = sum_{psi in S_n} operatorname{sgn}(psi circ tau) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j}.$$
But, since composing a permutation with a transposition changes its sign, we thus have,
$$det(beta_{i,j}) = -sum_{psi in S_n} operatorname{sgn}(psi) prod_{j in lbrace 1, ldots, n rbrace} alpha_{psi(j),j} = det(alpha_{i,j}).$$
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
answered Jul 4 '17 at 17:32
Theo BenditTheo Bendit
21k12355
21k12355
$begingroup$
Do you mean to say switch the two columns? My indexing convention is that if I have $a_{ij}$, then $i$ denotes the row, and $j$ the column.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:42
1
$begingroup$
Also, you have apparently shown what I have, which is that switching two rows (or columns) amounts to changing the sign. I am trying to show that if two neighboring columns (or rows) are identical, then the determinant is zero.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:44
add a comment |
$begingroup$
Do you mean to say switch the two columns? My indexing convention is that if I have $a_{ij}$, then $i$ denotes the row, and $j$ the column.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:42
1
$begingroup$
Also, you have apparently shown what I have, which is that switching two rows (or columns) amounts to changing the sign. I am trying to show that if two neighboring columns (or rows) are identical, then the determinant is zero.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:44
$begingroup$
Do you mean to say switch the two columns? My indexing convention is that if I have $a_{ij}$, then $i$ denotes the row, and $j$ the column.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:42
$begingroup$
Do you mean to say switch the two columns? My indexing convention is that if I have $a_{ij}$, then $i$ denotes the row, and $j$ the column.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:42
1
1
$begingroup$
Also, you have apparently shown what I have, which is that switching two rows (or columns) amounts to changing the sign. I am trying to show that if two neighboring columns (or rows) are identical, then the determinant is zero.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:44
$begingroup$
Also, you have apparently shown what I have, which is that switching two rows (or columns) amounts to changing the sign. I am trying to show that if two neighboring columns (or rows) are identical, then the determinant is zero.
$endgroup$
– Alex Ortiz
Jul 4 '17 at 17:44
add a comment |
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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
