2
$\begingroup$

In Section 11.4 of Dummit and Foote, they introduce a determinant function $\det$ on the ring of $n\times n$ matrices over a commutative ring $R$ with $1$ as

  1. Any $n$-multilinear alternating form, where the $n$-tuples are the $n$ columns of the matrices in $M_{n\times n}(R)$, and

  2. $\det(I) = 1$ where $I$ is the $n\times n$ identity matrix.

They then define a function $$ \det(\alpha_{ij}) = \sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\alpha_{\sigma(1)1}\dotsb\alpha_{\sigma(n)n} $$ 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.

$\endgroup$
2
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $\sum\limits_{\sigma \in S_n} \ldots$ 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 $k \in \left\{1,2,\ldots,n-1\right\}$. 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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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, $$\beta_{i,j} = \left\lbrace \begin{array}{ccc} \alpha_{i, p} &\mathrm{if} & j = q \\ \alpha_{i, q} &\mathrm{if} & j = p \\ \alpha_{i, j} & & \mathrm{otherwise}\end{array} \right.$$ 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, $$\det(\beta_{i,j}) = -\sum_{\psi \in S_n} \operatorname{sgn}(\psi) \prod_{j \in \lbrace 1, \ldots, n \rbrace} \alpha_{\psi(j),j} = \det(\alpha_{i,j}).$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.