A sum of determinants Let n be a positive integer, $n \ge 2$ and $ A \in \mathcal{M}_{n}(\mathbb{C}) $. $M$ is the set of matrices which are obtained by swapping exactly $2$ elements of the same row or column of $A$. Calculate $ \sum_{X \in M} det X$, depending on A.
I saw that for $n = 2$ the sum is $0$ and my intuition(verified on small numbers) would be that for $n > 2$, the sum should be equal to $2n\binom{n-1}{2}detA$ (for $n=3$ the sum is $6detA$). The reason is that I think that for the products that don't appear by expanding in the original determinant A, we can find their opposite in other matrices from $M$. Now, by switching elements and counting, we can see that the products we find by expanding $detA$ appear $2n\binom{n-1}{2}$ times. However, i don't have a clear proof of the fact that other products find their opposite.
By product, i was referring to something of the type of $sgn(\sigma)a_{1\sigma(1)}...a_{n\sigma(n)}$ that show up in all the matrices from $M$, since $detA = \sum_{\sigma \in S(n)} sgn(\sigma)a_{1\sigma(1)}...a_{n\sigma(n)}$ and we consider $A = (a_{ij})_{n×n}$.
Can you help me on this one?
 A: You can define a function $F(v_1,..., v_n) = \sum_{X\in M} \det(X)$, where the $M$ is built out of the matrix whose columns are the vectors $v_1,..., v_n$.
This function is clearly multilinear as the operations of scaling or adding two vectors in a single row or column commute with swapping a specific two elements in a specific row or column.
We want to see that this is alternating, so suppose that $v_i = v_j$ and for concreteness let us suppose $v_i = v_j = (a_1, a_2,..., a_n)^T$. When you consider the value of $F(v_1,..., v_n)$ in this case several things can happen:

*

*If the swap occurs and no term from the column $i$ or $j$ is moved then the resulting matrix has $0$ determinant since it has repeated columns.

*If the swaps changes the two elements in some row that correspond to coumns $i$ and $j$ then again we end up with something that has repeated columns and so determinant $0$.

This tells us that the changes either occur within column $i$ or $j$, or it swaps an element of some row involving exactly one of these columns. We will see in either case we end up having cancelations in the value of $F$.
If the swap occurs within column $i$, swapping to elements, say WLOG the ones of the first two rows, then there is also another matrix in $M$ corresponding to making this same swap but instead in the column $j$. Consider then the vector $w= (a_1 + a_2, a_1 + a_2, 2a_3,..., 2a_n)^T$ and realize that
$\det(v_1,..., v_{i-1}, w, v_{i + 1},..., v_{j-1}, w, v_{j + 1},..., v_n) = 0$
because it has repeated columns. On the other hand, by the linearity of determinant we get that this same determinant is the same as distributing the sums. Concretely if $w_1 = (a_2, a_1, a_3,..., a_n)^T then the determinant is
$0 = \det(v_1,..., v_{i-1}, w_1, v_{i + 1},..., v_{j-1}, w_1, v_{j + 1},..., v_n)\\+ \det(v_1,..., v_{i-1}, w_1, v_{i + 1},..., v_{j-1}, v_2, v_{j + 1},..., v_n)\\ + \det(v_1,..., v_{i-1}, v_1, v_{i + 1},..., v_{j-1}, w_1, v_{j + 1},..., v_n)\\+ \det(v_1,..., v_{i-1}, w_1, v_{i + 1},..., v_{j-1}, w_1, v_{j + 1},..., v_n)\\=
\det(v_1,..., v_{i-1}, w_1, v_{i + 1},..., v_{j-1}, v_2, v_{j + 1},..., v_n)\\ + \det(v_1,..., v_{i-1}, v_1, v_{i + 1},..., v_{j-1}, w_1, v_{j + 1},..., v_n)$
and this last expression are two terms corresponding to the aforementioned swaps, so they cancel out.[More succintly, the two resulting matrices have two columns exchanged.]
Finally, if the swap involves changing two elements of a row and exactly one of them is ins row $i$, then of course we have the corresponding swap but changing instead the element in the column $j$. The two obtained matrices, after these swaps, are almost the same, except that they have column $i$ and $j$ are interchanged. This implies the determinant of one is minus the determinant of the other, and so when you do the sum in $F$ they also cancel out. [These cancelling outs are the ones you say you expect, but in the expansion version of the determinant is hard to keep track of them.]
These are all possibilities, so we get $F$ is alternating. Since the determinant generated the multilinear alternating forms we get $F(X) = c\det(X)$ for some contant $c$. To evaluate the constant we plug in for $X$ the identity matrix (i.e. the standard vectors).
Notice that no swap can involve two ones, since they appear in different columns and different rows so, either we exchange two zeroes or one $1$ with a $0$. In the latter case, we create a matrix that has a row or column consisting of only $0$'s so its determinant is 0.
This implies that $F(I)$ is exactly the number of ways of swapping two zeroes that are in the same row or the same column. There are $2n$ of picking the row/column, and $\binom{n-1}{2}$ ways of picking the two zeroes. Furhtermore, each obtained matrix is still the identity so its determinant is $1$. We conclude $c = F(I) = 2n\binom{n - 1}{2}$, that is
$F(X) = 2n\binom{n - 1}{2}\det(X)$
as claimed.
