What is the expectation of the rank of a matrix with a 1 at each column? Say a random square matrix $A\in\mathbb{R}^{n\times n}$, each column of $A$ has exactly one nonzero element being 1, i.e. each column looks like $e_i=\{0,\dots,1,\dots,0\}^\top$. Say for each column, the position of 1 is i.i.d. draw from the uniform distribution over $\{1,2,\dots,n\}$. What will the distribution of $rank(A)$ look like? What is its expectation? Will it concentrate near its expectation? 
 A: The rows of $A$ are pairwise orthogonal, so the rank of $A$ is the number of nonzero rows. Each row is nonzero with probability $1-(1-1/n)^n$, so the expected rank is $n(1-(1-1/n)^n)\approx n(1-1/e)\approx0.632\times n$ for large $n$. For small $n$, we have the values $1,\frac32,\frac{19}{9},\frac{175}{64},\ldots$.
(Besides the expectation, I'm not sure what the rest of the distribution looks like.)
A: The rank is the number of rows containing at least one $1$.  We will use the method of indicator variables to find the mean and variance of the rank.  It turns out that the distribution does concentrate near the mean.
Let $$I_i =
\begin{cases}
1 \qquad \text{if row } i \text{ contains at least one 1} \\
0 \qquad \text{otherwise}
\end{cases}$$
for $i = 1,2,3, \dots ,n$, and let 
$$X = \sum_{i=1}^n I_i$$
so $X$ is the rank of the matrix.  We have
$$P(I_i = 0) = \left( \frac{n-1}{n} \right)^n = (1-1/n)^n$$
so $P(I_i = 1) = 1- (1-1/n)^n$, and the mean of $X$ is
$$\mu = E(X) = E \left( \sum_{i=0}^n I_i \right) = \sum_{i=0}^n E( I_i) = n [ 1- (1-1/n)^n]$$
We know $(1-1/n)^n \to e^{-1}$ as $n \to \infty$, so $\mu \sim (1-e^{-1})n$.
To find the variance of $X$, we need to first find $E(\sum_{i<j} I_i I_j)$.  So notice that
$$\begin{align}
P(I_i I_j = 0) &= P(I_i = 0) + P(I_j = 0) - P((I_i = 0) \cap (I_j=0)) \\
&= 2 (1-1/n)^n - (1-2/n)^n
\end{align}$$
for $i < j$, so $P(I_i I_j = 1) = 1 -2 (1-1/n)^n + (1-2/n)^n$.  Therefore
$$E \left( \sum_{i<j} I_i I_j \right) = \sum_{i<j} E(I_i I_j) = \frac{n(n-1)}{2} [ 1 -2 (1-1/n)^n + (1-2/n)^n ] $$
Because of the identity
$$\left( \sum_{i=0}^n I_i \right )^2 = \sum_{i=0}^n I_i^2 + 2 \sum_{i<j}I_i I_j = \sum_{i=0}^n I_i + 2 \sum_{i<j}I_i I_j$$
we have
$$X^2 = X + 2 \sum_{i<j}I_i I_j$$
so
$$\begin{align}
E(X^2) &= E(X) +  2 \sum_{i<j} E(I_i I_j) \\
&= n [ 1- (1-1/n)^n] + n(n-1) [ 1 -2 (1-1/n)^n + (1-2/n)^n ]
\end{align}$$
So the variance of $X$ is
$$\begin{align}
\sigma^2 &= E(X^2) - (E(X))^2 \\
&= n(1-1/n)^n - n^2(1-1/n)^{2n} + n(n-1)(1-2/n)^n
\end{align}$$
after some simplification.  It turns out (see below) that $\sigma^2 \sim (e^{-1} - 2 e^{-2}) n$, so
$$\frac{\sigma}{\mu} \sim \frac{\sqrt{e^{-1}-2 e^{-2}}}{1-e^{-1}} \frac{1}{\sqrt{n}}$$
and the distribution does concentrate near the mean for large $n$.

The asymptotic behavior of $\sigma^2$ for large $n$:
From above, we have
$$\begin{align}
\frac{\sigma^2}{n} &= (1-1/n)^n - n (1-1/n)^{2n} + (n-1)(1-2/n)^n \\
&= (1-1/n)^n - (1-2/n)^n+ n[(1-2/n)^n-(1-1/n)^{2n}] 
\end{align}$$
so
$$\lim_{n \to \infty} \frac{\sigma^2}{n} = e^{-1} - e^{-2} + \lim_{n \to \infty} n[(1-2/n)^n -(1-1/n)^{2n}] \tag{*}$$
To evaluate the last limit above, notice that
$$\lim_{n \to \infty} n[(1-2/n)^n -(1-1/n)^{2n}] = \lim_{x \to 0} \frac{1}{x} [(1-2x)^{1/x} - (1-x)^{2/x}] $$
and
$$\begin{align}
(1-2x)^{1/x} &= \exp( (1/x) \ln(1-2x)) \\
&= \exp((1/x)(-2x -2x^2 +o(x^3)) \\
&= \exp(-2 -2x +o(x^2)) \\
&= e^{-2} \exp(-2x +o(x^2)) \\
&= e^{-2} (1-2x) + o(x^2)
\end{align}$$
Similarly,
$$\begin{align}
(1-x)^{2/x} &= \exp((2/x) \ln(1-x) \\
&= \exp((2/x) (-x -(1/2)x^2 +o(x^3)) \\
&= \exp(-2-x+o(x^2)) \\
&= e^{-2} \exp(-x +o(x^2)) \\
&= e^{-2} (1-x) + o(x^2)
\end{align}$$
So
$$\begin{align}
\lim_{x \to 0} \frac{1}{x} [(1-2x)^{1/x} - (1-x)^{2/x}] &= \lim_{x \to 0} \frac{1}{x} [e^{-2}(1-2x) +o(x^2) - e^{-2}(1-x) + o(x^2)] \\
&= \lim_{x \to 0} (-e^{-2} + o(x)) \\
&= -e^{-2}
\end{align}$$
Substituting into (*), we have
$$\lim_{n \to \infty} \frac{\sigma^2}{n} = e^{-1} - e^{-2} - e^{-2} = e^{-1} -2 e^{-2}$$
so $\sigma^2 \sim (e^{-1} - 2 e^{-2}) n$, as was claimed above.
