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.


It might be an overkill, but here is my take on this question:

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.

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  • $\begingroup$ Thanks for the insight! I didn't know that are rules for computing the rank of a matrix arising from a kronecker product! $\endgroup$ – Tomas Jorovic Jan 24 at 19:37

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