Why is the determinant of a uniformly distributed matrix less than the determinant of a normally distributed one? I was solving Problem set 6 in Gilbert Strang's famous MIT OCW Linear Algebra course (link) and I came across the following problem:
What is the typical determinant (exper­imentally) of rand(n) and randn(n) for $n=50,100,200,400$?
The solution shows experimentally that the determinants with $rand()$ are smaller than the determinants with $randn()$. For rand(), the solution states that rand(50) is around $105$, rand(100) is around $1025$, rand(200) is around $1079$, and rand(400) is around $10219$. For $randn()$, the solution states that randn(50) is around $1031$, randn(100) is around $1078$, randn(200) is around $10186$,and randn(400) is just too big for matlab.
However, intuitively I was expecting the opposite because I think normally distributed numbers would have a higher chance of being close to $0$ than uniformly distributed numbers, thus making it more likely to have a smaller determinant for the randn() case. I am not able to understand what I am missing here. Could someone please give an intuitive or a mathematical reason behind the experimental results shown in the above problem?
 A: The question is a bit ambiguous. Since the matrix entries are i.i.d., for any matrix sample $A$, if $B$ is the matrix obtained by swapping the first two columns of $A$, then $A$ and $B$ should have the same likelihood of occurrence. It follows that the mean determinants for rand(n) and randn(n) are always zero, unless $n=1$.
So, I will consider the absolute determinant instead. I don't know how to analyse the median or mode of the absolute determinant, but I can analyse the mean. Let $A$ be a random sample of randn(n). Denote the first column of $A$ by $\mathbf a$. Let $Q_1$ be any orthogonal matrix whose first column points to the same direction as $\mathbf a$. Then
$$
B=Q_1^TA=\pmatrix{\|\mathbf a\|&b_{12}&\cdots&b_{1n}\\ 0&b_{22}&\cdots&b_{2n}\\ \vdots&\vdots&&\vdots\\ 0&b_{n2}&\cdots&b_{nn}}.
$$
Since the entries of $A$ are i.i.d. standard normal, $E(\|\mathbf a\|)=\sqrt{n}$. Now let $\mathbf b=(b_{22},b_{32},\ldots,b_{n2})^T\in\mathbb R^{n-1}$. As the choice of $Q$ depends only on $\mathbf a$, the elements in the second to the last columns of $B$ are still i.i.d. standard normal. So, if we pick an orthogonal matrix $Q_2\in O_{n-1}(\mathbb R)$  whose first column points  to the same direction as $\mathbf b$, then
$$
C=\pmatrix{1\\ &Q_2^T}B=\pmatrix{\|\mathbf a\|&b_{12}&\cdots&\cdots&b_{1n}\\ 0&\|\mathbf b\|&c_{23}&\cdots&c_{2n}\\
0&0&c_{33}&\cdots&c_{3n}\\
\vdots&\vdots&&\vdots\\
0&0&c_{n2}&\cdots&c_{nn}}
$$
where the $c_{ij}$s are i.i.d. normal and are independent from $\|\mathbf a\|$ and $\|\mathbf b\|$. Continue in this manner, we finally obtain $E(|\det(A)|)=E(\|\mathbf a\|\|\mathbf b\|\cdots)=E(\|\mathbf a\|)E(\|\mathbf b\|)\cdots=\sqrt{n!}$, which grows like $\sqrt{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}$, by Stirling's formula.
Our previous argument is actually adapted from a proof of Hadamard's determinant inequality, which says that $|\det(A)|\le\prod_{j=1}^n\|\mathbf a_j\|$ (where $\mathbf a_j$ denotes the $j$-th column of $A$), because when $A=QR$ is a QR factorisation, we have
$$
|\det(A)|=|\det(R)|=\prod_j|r_{jj}|\underset{(\ast)}{\color{red}{\le}}\prod_j\|\mathbf r_j\|=\prod_j\|\mathbf a_j\|.
$$
What if the elements of $A$ is sampled from rand(n) instead? By Hadamard's inequality, we have
$$
E(|\det(A)|)\le\prod_jE(\|\mathbf a_j\|)=\sqrt{\left(\frac{n}{3}\right)^n}.
$$
So, even if we assume that the upper bound $(\ast)$ we marked in red above is always tight, the mean absolute determinant still grows slower than when $A\sim$ randn(n).
