[AntonioLao]

It is multiples of the defined unit vectors. In a 3-space Cartesian system, the unit vectors are symbolized by i, j, and k , all of unit length. The length of a vector V=a i+b j+c k is defined as the absolute magnitude |V| that is the square root of a²+b²+c². Furthermore, these 3 unit vectors can each be represented by 3 by 1 column or 1 by 3 row matrices (1, 0, 0), (0, 1, 0), and (0, 0, 1). All these matrices are linearly independent and they serve as a basis for spanning any given vector in a 3 dimensions vector space where a, b, and c are all real numbers.

However, if the zeros are replaced by -1 and 1 then the basis becomes (1, -1, -1), (-1, 1, -1), and (-1,-1, 1). Their scalar product with -1 produced a new basis (-1, 1, 1), (1, -1, 1), and (1, 1, -1) which are also linearly independent with themselves but not with the other basis. This linear dependence is determined by scalar product of matrix addition such that there exist scalars: a, b, c, d, e, f, not all zeros but still a(1, -1, -1)+b(-1, 1, -1)+c(-1, -1, 1)+d(-1, 1, 1)+e(1, -1, 1)+f(1, 1, -1)=0.

On the other hand, the matrix multiplications of these 2 basis produced single elements or squares of 3 by 3 matrices. Although these can be singular as well as nonsingular depending on the arrangements of -1 and 1, nonetheless for products of basis there are no exceptions to the rule of singularity. At the infinitesimal domain, quantum vectors have lengths of 1, √2, and √3 within a unit cube. On the other hand, spinors and twistors all have length zeros or imaginary.