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In the book of Linear Algebra by Werner Greub, at page $111$, question $2$,

Prove that the determinant of the $n \times n$ matrix $A = (\alpha_v^u)$, where $$\alpha_v^u = 1- \delta_v^u,$$ is equal to $(n-1)(-1)^{n-1}$.

If we consider $A$ as the matrix of the map $\phi : E \to E, (dim E = n)$, with respect to the basis $e_v$ , we can say that

$$\phi (e_v) = (\sum_u e_u) - e_v$$, so by definition

$$\Delta_\phi(e_1,..., e_n)= \Delta(\phi(e_1), ..., \phi (e_n)) = \Delta (\sum_{u \not = 1} e_u, ..., \sum_{u \not = n} e_u) = det \phi$$, where $\Delta$ is a nonzero determinant function and $\Delta (e_1, ..., e_n) = 1$, and from that the the only contribution will come from the derangement of $e_v$s, but the number of derangements is huge compare to the $(n-1)(-1)^{n-1}$, so how can continue from that ?

I would appreciate help.

Note: In here, there is a answer to my question, and its link is given, but I would specifically like to learn how to continue from the point that I have arrived because, for example, if I tried to solve this very same question after a month, I will again use a method similar to this one.

Edit:

I'm particularly looking for a proof that continues from where I left.

Note 2:

After 3 months that I have first faced with this question, I have tried to solve it again, and used the same method as the my first attempt above, and stuck in a similar point in the answer given to this question.

darij grinberg
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Our
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  • Are you trying to solve the problem by showing that the number of even derangements minus the number of odd derangements is $\left(n-1\right) \left(-1\right)^{n-1}$ ? I'm not sure whether this is any easier than the original problem. Exercise 9 (c) in Mark Wildon's An involutive introduction to symmetric functions does this using symmetric functions theory, but I somehow doubt that you're looking for that proof :) – darij grinberg Oct 01 '17 at 06:09
  • @darijgrinberg The continuation might not be easy to understand, but I least I should know that. – Our Oct 01 '17 at 06:14
  • @darijgrinberg I mean I have stuck in the same point in the below answer, so I thought I should try to continue from my starting. – Our Oct 01 '17 at 06:15

1 Answers1

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We can use multilinearity: let $v_i:=\sum_{j=1,j\neq i}^ne_j $. We want to compute $\Delta:=\det\left(v_1,\dots,v_n\right)$. Observe that $$v_i =\sum_{j=1} ^ne_j-e_i ,$$ hence $$\sum_{i=1}^nv_i=n\sum_{j=1} ^ne_j - \sum_{i=1}^ne_i=\left(n-1\right)\sum_{i=1}^ne_i.$$ Consequently replacing the $n$-th element of $\left(v_1,\dots,v_n\right)$ by $\sum_{i=1}^nv_i$, we get
$$\Delta=\det\left(v_1,\dots,v_{n-1},\left(n-1\right)\sum_{i=1}^ne_i \right)=\left(n-1\right)\det\left(v_1,\dots,v_{n-1},\sum_{i=1}^ne_i \right).$$ Now, replace $v_i$ by $v_i-\sum_{l=1}^ne_l$ to get that $$\Delta=\left(n-1\right)\det\left(v_1-\sum_{l=1}^ne_l,\dots,v_{n-1}-\sum_{l=1}^ne_l,\sum_{i=1}^ne_i \right).$$ Since $v_u-\sum_{i=1}^ne_i=-e_u$, we derive that $$\Delta=\left(n-1\right)\det\left(-e_1,\dots,-e_{n-1} ,\sum_{i=1}^ne_i \right),$$ giving the wanted result.

Davide Giraudo
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  • So, after "Consequently", shouldn't it be like $ v_n = \sum_{i=1}^{n-1} e_i $, so you cannot do that replacement of yours. – Our Jul 07 '17 at 11:08
  • I can, since for any $i\in\left{1,\dots,n-1\right}$, the determinant $\det(v_1,\dots,v_{n-1},v_i )$ is zero. – Davide Giraudo Jul 07 '17 at 11:18
  • Ok, what about replacing $v_i$ by $v_i-\sum_{l=1}^ne_l$ ? I mean $\sum_{l=1}^ne_l = v_i + e_i$ – Our Jul 07 '17 at 11:21
  • It is just the use of multilinearity of the determinant (when two columns are the same, it is zero). – Davide Giraudo Jul 07 '17 at 11:31
  • Hi again @DavideGiraudo, I revisited your answer, and again stuck at the point "Now, replace $v_i$ by $v_i-\sum_{l=1}^ne_l$", can you edit your answer, so that the answer can become clear for every one. – Our Sep 30 '17 at 16:52
  • Look, last time it took my 16 days to figure out the thing that I was missing, so I think it would be really beneficial if you can explain that point more clearly. – Our Sep 30 '17 at 16:53