Consider a full rank matrix $A\in \mathbb{R}^{m\times n}$ with $n\geq m$, let $a_j$ be its $j$-th column then $a_j,j=1,\ldots,n$ span the space $\mathbb{R}^m$, meaning that they are a frame basis of $\mathbb{R}^m$. My question is: are vectors $a_j \otimes a_j,j=1,\ldots,n$ linearly independent?
Edit: I found a simple counter-example: just let $A = [I_m 0]$ then $a_j\otimes a_j,j=1,\ldots,n$ will only have $m$ nonzero vectors so it can not be linearly independent.
Question: Since I now have a bad example, I am very curious what conditions matrix $A$ should satisfy to make $\{a_j\otimes a_j\}$ independent. Should we assume the frame basis $a_j$ is spread even, i.e. they form a Parseval frame basis? Would it be sufficient?
Edit2: user10354138's comments are very helpful and my initial question is not a good question. Let me rephrase my question below:
Consider a full rank matrix $A\in \mathbb{R}^{m\times n}$ with $n\geq m$ and $n \leq m^2$, let $a_j$ be its $j$-th column then $a_j,j=1,\ldots,n$ span the space $\mathbb{R}^m$, meaning that they are a frame basis of $\mathbb{R}^m$. My question is: what assumptions shoud $a_j$ satisfy to guarantee that vectors $a_j \otimes a_j,j=1,\ldots,n$ are linearly independent?
Because of my counter-example above, my guess is $a_j$ should be a nice frame, i.e. a Parseval frame basis.