Take two vectors $a,b\in \mathbb{R}^{12}$ and create a 12x12 matrix $C$ so that $c_{ij}=(a_i-b_j)^2$. What are the possible values of rank(C)?
Not sure how to proceed. One idea I had was to express C as a product of two other matrices and then I might put together an estimate of some sort using $rank(AB)\leq min \{ rank(A),rank(B) \}$. Though having tested it out on a couple of concrete 3x3 matrices I found they always had full rank. So maybe that's the answer?
Either way, appreciate any help.
I believe the rank is no greater than $3$. Notice that by expanding out $c_{ij}=(a_i-b_j)^2 = a_i^2 + b_j^2 - 2a_ib_j$, we can write
$$ C = (a^2) \boldsymbol{1}^T + \boldsymbol{1} (b^2)^T - 2ab^T$$
Where $\boldsymbol{1}$ is the vector of all ones and $a^2$ is the vector whose entries are the squared entries of $a$. Hence there are at most three linearly independent columns: $a,a^2,$ and $\boldsymbol{1}$. When $a,a^2$, and $\boldsymbol{1}$ are all linearly independent (and analogously for the rows $b,b^2,\boldsymbol{1}$) the rank is 3. We can also obtain rank $2,1,$ or $0$. Choosing $a=a^2\neq \boldsymbol{1}$ gives rank $2$. Choosing $\boldsymbol{1} = a = a^2$ gives rank $1$. And $a=a^2=b=b^2=0$ trivially gives rank zero.