This will take me some time to explain. Let $n geq 2$ be a fixed integer. Let $p_i(z)$, for $i = 1,ldots,n$ be $n$ nonzero complex polynomials of degree at most $n-1$. I am interested in reformulating whether or not the $n$ complex polynomials are linearly independent over $mathbb{C}$, using the so called “squaring map”, which is familiar to people working with spinors. It is a quadratic map which maps spinors to vectors. I will define everything.

For $i = 1, ldots, n$, write

$$ p_i(z) = prod_{j neq i} L_{ij}(z) $$

where $j$ runs over the values from $1$ to $n$ that are different from $i$, and each $L_{ij}(z)$ is a (nonzero) complex polynomial of degree at most $1$, so that

$$L_{ij}(z) = a_{ij} z + b_{ij}.$$

Of course, for a fixed $i$, the $L_{ij}$ are not uniquely determined. For instance these linear factors can be permuted, or you could for instance scale them differently. In the end, our construction will be independent of such ambiguities.

Given a nonzero polynomial

$$L(z) = az + b$$

of degree at most $1$, one can form a nonzero element of $psi_L in mathbb{C}^2$ out of its coefficients, namely

$$psi_L = left( begin{array}{c} a \ b end{array} right).$$

We can now define the “squaring” $Sq(L)$ of the polynomial $L(z)$ to be

$$ Sq(L) := psi_L psi_L^* = left( begin{array}{cc} |a|^2 & a bar{b} \ bar{a} b & |b|^2 end{array} right). $$

We also define the “squaring” of a nonzero complex polynomial of degree at most $n-1$ to be

the symmetric tensor product (over $mathbb{R}$) of the squarings of its linear factors. So for instance,

$$ Sq(p_i) := odot_{j neq i} Sq(L_{ij}). $$

I should say what I mean by the symmetric tensor product. Each $Sq(L_{ij})$ has an index up and an index down, with each of these indices taking values in ${1,2}$. In order to define $Sq(p_i)$, first form the outer product (or tensor product if you prefer) of all the $Sq(L_{ij})$, for $j neq i$, then completely symmetrize all the indices up together, and completely symmetrize all the indices down together. Since each index can only take on $2$ possible values, this will result in a tensor, which can be viewed simply as an $n times n$ matrix, one index up (resp. down) which corresponds to the various values of the symmetrized indices up (resp. down).

One can check that $Sq(p_i)$ is indeed well defined, despite the ambiguities in the definitions of the factors $L_{ij}$.

Define the following “sum of squares”:

$$ S := sum_{i=1}^n Sq(p_i),$$

which can be thought of as an $n times n$ matrix, by the remarks above.

I conjecture that the $p_i$, for $i = 1, ldots, n$, are linearly independent over $mathbb{C}$ iff their “sum of squares” $S$ is nonsingular.

For $n = 2$, this is straightforward. I did some numerical simulations for $n = 3$, which seem to confirm my conjecture. However, I did not do any numerical simulations for $n > 3$. So it could be false for $n = 4$ perhaps. I am not sure yet.

If someone has any comments and/or, better, knows how to prove/disprove my conjecture, then please share your knowledge.