# Rank of a matrix formed by pairwise products of columns of another matrix

Let $$A$$ be a full rank matrix such that $$A = [a_1, a_2, a_3]$$ are its columns. Suppose that the number of rows of $$A$$ is larger than 6. Consider $$B$$ such that $$B=[a_1^2, a_1a_2, a_1a_3 ,a_2^2, a_2a_3, a_3^2]$$ in which $$a_i a_j$$ denotes the column vector formed by element-wise multiplication of the column vectors $$a_i$$ and $$a_j$$. For example, if $$a_i = [1,2,3]^T$$ and $$a_j = [2,4,6]^T$$ then $$a_i a_j=[2,8,18]^T$$.

Are there known results to determine under what conditions $$B$$ has full rank? I am of course considering the general case when $$A$$ has more than 3 columns.

Just say $$A\in \mathbb{R}^{m \times n}$$, where $$m\geq \frac{1}{2}\cdot n(n+1)$$, i.e. it has enough columns to let $$B$$ be at least square. Let $$C:=A\otimes A$$ denote the Kronecker product of $$A$$ with itself. The gives $$C\in \mathbb{R}^{nm\times nm}$$ which is a very large matrix, that contains all the entries of $$B$$ somewhere in it.
And this is the point: $$C$$ always has full rank, if $$A$$ is of full rank. Your matrix $$B$$ is hidden somewhere and can be found by deleting some rows (e.g. the second row, where the first entries of any $$a_i$$ is multiplied with the second entry of any $$a_j$$) and some columns (as there are $$a_1 \odot a_2$$ and $$a_2\odot a_1$$ in $$C$$ but not $$B$$).
If you delete these rows from $$C$$ this will not become rank-deficient as can be inferred by the SVD of $$C$$. By deleting rows and colums from the elft and singular vectors, you never form a zero-singular value.