TITLE: Expected Value of a Determinant
QUESTION [16 upvotes]: Suppose that I construct an $n \times n$ matrix $A$ such that each entry of $A$ is a random integer in the range $[1, \, n]$. I'd like to calculate the expected value of $\det(A)$.
My conjecture is that the answer is zero, though I could very well be incorrect. Running some numerical experiments with different values for $n$ and a large number of trials, it seems that $\mathbb{E}[\det(A)]$ is normally in the range $[0.25, \, 0.7]$, so I'm starting to lose faith in my intuition that it is zero.
Could anyone lend some advice on how to approach this problem and what strategies I may want to consider applying?

REPLY [2 votes]: Let $A_n$ be an $n \times n$ matrix.
The case of a $1 \times 1$ matrix is a little different than the general case. Since the determinant is just the single entry itself, it follows that since the entry must be one, the determinant is also one.
For $n \geq 2$, you can show that the answer is zero using induction. Consider the case $n=2$ and the corresponding matrix $A_2$. It's determinant is
$$
\det(A_2) = \left| \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \right| = ad - bc.
$$
In other words,
$$
\mathbb{E}[\det(A_2)] = \mathbb{E}[ad-bc] = \mathbb{E}[ad] - \mathbb{E}[bc] = \mathbb{E}[a]\mathbb{E}[d] - \mathbb{E}[b]\mathbb{E}[c]
$$
where the second equality comes from the linearity of the expected value and the second comes from the fact that $a$, $b$, $c$, and $d$ are independent random variables. It's not hard to see that $\mathbb{E}[X] = (2+1)/2 = 3/2$ for $X \in \{ a, \, b, \, c, \, d \}$. Thus,
$$
\mathbb{E}[\det(A_2)] = \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 = 0.
$$
Next assume that $\det(A_n) = 0$, where $A_n$ is an $n \times n$ matrix.
For the inductive step, note that $\det(A_{n+1})$ can be written as
$$
\det(A_{n+1}) = \lambda_1\det\left(A_{n}^{(1)}\right) - \lambda_2\det\left(A_{n}^{(2)}\right) + \lambda_3\det\left(A_{n}^{(3)}\right) - \cdots \pm \lambda_{n+1}\det\left(A_{n}^{(n+1)}\right)
$$
by using the usual method to expand the determinant across the top row (or any row/column, really) of the $(n+1) \times (n+1)$ matrix (note that I used the superscripts simply to index the $n \times n$ matrices). Since $\det\left(A_n^k\right) = 0$ for $k = 1, \, 2, \dots, \, n+1$, the determinant is zero for $n \geq 2$, by induction.
Putting these results together, we have
$$
\mathbb{E}[\det(A_n)] = \begin{cases}
1, & n = 1 \\
0, & n \geq 2.
\end{cases}
$$
Note that you can easily extend this result to show that if the entries come from any bounded set, the determinant must be zero for $n \geq 2$.