Average Determinant

Question

Let $S$ be the set of all $n \times n$ real matrices ($n$ > 1) with entries from a finite set $M$. Can we say that the average value of determinants of matrices in $S$ is always zero?

See that $S$ is closed under row interchange. So, we have matrices with equal determinant but of opposite sign. I think, we have to show that there is a bijective map between the subset of matrices with positive determinant and subset of matrices with negative determinant. Is that enough? Observe that the problem is independent of nature of the entry set $M$.

Answer 1

I think, we have to show that there is a bijective map between the subset of matrices with positive determinant and subset of matrices with negative determinant. Is that enough?

The idea is correct, but the above formulation is not. There is a bijection between the sets $\{1\}$ and $\{-4\}$, one containing only positve numbers, the other only negative numbers, still the average of all numbers is not $0$.

The correct formulation of the idea would be to try to find a bijective map between the subset of matrices with positive determinant and subset of matrices with negative determinant, where the corresponding matrices have determinants of the same absolute value (but opposite signs).

As you noted, interchanging 2 fixed rows (or columns) is exactly such a map. It changes the sign of the determinant, but keeps the absolute value. Depending on the rigor needed/wanted for the proof, you should think about why this is really a bijection! It's not hard, but sometimes in such proofs that is not so easy, and sometimes what looks like a bijection isn't one.

Answer 2

Swapping rows $1$ and $2$ gives you an involution of $S$; in more words, a function $i:S\to S$ such that $i\circ i=\text{id}_S$. Since $i$ is its own left and right inverse, it is a bijection and its own inverse.

We have a function $\det:S\to\mathbb R$ which takes a matrix to its determinant. A standard fact (sometimes part of the definition) of the determinant is that interchanging rows negates the determinant. In symbols: $\det\circ i=-1\circ\det$, where I mean $-1:\mathbb R\to \mathbb R$ to be the multiplication by $-1$ function.

In order to show that the average of the determinants is $0$, it is enough to show that the sum of the determinants is $0$: $$\sum_{s\in S} \det(s)= 0.$$ Write $S$ as the disjoint union of subsets $S_+, S_0, S_-$ consisting of the elements with positive, zero, and negative determinant respectively. Then we can expand the sum over this partition: $$\sum_{s\in S_+} \det(s) + \sum_{s\in S_0} \det(s) + \sum_{s\in S_-}\det(s).$$ By construction, the middle sum is $0$, since each term of the sum is $0$. Each term in the left sum is positive, and each term in the right sum is negative. It would be great if we could pair them off.

Well, for $s\in S_+$, we have $\det(i(s)) = -\det(s) < 0$, so $i(s)\in S_-$. Similarly, $i$ sends $S_-$ to $S_+$. We have two functions $i_1:S_+\to S_-$ and $i_2:S_-\to S_+$ which are the respective restrictions of $i$. Furthermore, since $i$ is an involution, we have $i_1\circ i_2 = \text{id}_{S_-}$ and $i_2\circ i_1 = \text{id}_{S_+}$. In particular, $i_1$ and $i_2$ are inverse functions, so they are bijections. This means that each $s\in S_-$ can be uniquely written as $i(s')$ for $s'\in S_+$. Returning to our sum of determinants: $$\sum_{s\in S_+} \det(s) + \sum_{s\in S_-} \det(s)$$ $$= \sum_{s\in S_+} \det(s) + \sum_{s'\in S_+} \det(i(s'))$$ $$\sum_{s\in S_+} \det(s) - \sum_{s'\in S_+} \det(s') = 0.$$