Let $A$ be a $m \times n$ random matrix with $m>n$ and $\operatorname{rank} A = n$ almost surely. Let $a$ be a random column vector of size $n \times 1$. Here $A$ is not necessarily independent of $a$. We have $$ \mathbb E[Aa] = 0 \in \mathbb R^m. $$
I wonder if $\mathbb E a=0$.
Let $a_i$ and $A_i$ be the $i$-th coordinate of $a$ and $i$-th column of $A$ respectively. Then $\mathbb E[Aa] = 0$ implies $$ \sum_{i=1}^n \mathbb E[a_i A_i] = 0 \in \mathbb R^m. $$
However, $a_i, A_i$ are not necessarily independent, so I can not use $\mathbb E[a_i A_i] = \mathbb E a_i \mathbb E A_i$. Could you elaborate if $\mathbb E a=0$?
No: one counterexample is to take the ordered pair $A,a$ to equal $\begin{pmatrix} 1&0 \\ 0&2 \\ 0&0 \end{pmatrix} , \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ with probability $\frac12$ and to equal $\begin{pmatrix} 2&0 \\ 0&1 \\ 0&0 \end{pmatrix} , \begin{pmatrix} -1 \\ -2 \end{pmatrix}$ with probability $\frac12$. Another is to take $a$ to be a nonzero constant vector and to let $A$ equal $\begin{pmatrix} 1&0 \\ 0&1 \\ 0&0 \end{pmatrix}$ or $\begin{pmatrix} -1&0 \\ 0&-1 \\ 0&0 \end{pmatrix}$ each with probability $\frac12$; note that in this example we even have $A$ and $a$ independent.
One analogy that might be helpful: the proposed statement doesn't even hold for real-number multiplication. It's easy to find random variables $X,Y$ (with $X\ne0$ almost surely) for which $\Bbb E[XY]=0$ but $\Bbb E[Y]\ne0$.