# Uniqueness of the basis for a tensor product of vector spaces.

Having in mind the following answered question:

Tensor Product of Spaces has Basis of Tensor Products

about bases for a tensor product of vector spaces I wonder if that is the only way to get a basis for that tensor other than the obvious. That is, the basis ${(e_1)}\otimes{(e_2)}$ for ${V_1}\otimes{V_2}$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2$.

In case it was not unique, would that mean a different tensor product between the corresponding matrices than the usual Kronecker product between matrices?

• Bases are never unique. – Paul Frost Jul 24 '18 at 8:15

To begin with, there are many basis for both $V_1$ and $V_2$ so there are several choices of basis on the form $(e_1) \otimes (e_2)$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2.$
Then, one can have "cross-bases". For example, assume that $V_1$ and $V_2$ are both 2-dimensional with bases $\{ e_1^{(1)}, e_1^{(2)} \}$ and $\{ e_2^{(1)}, e_2^{(2)} \}$ respectively. Then as a basis for $V_1 \otimes V_2$ you can take $$\{ e_1^{(1)} \otimes e_2^{(1)} + e_1^{(1)} \otimes e_2^{(1)}, e_1^{(1)} \otimes e_2^{(2)} + e_1^{(1)} \otimes e_2^{(2)}, \\ e_1^{(2)} \otimes e_2^{(1)} + e_1^{(2)} \otimes e_2^{(1)}, e_1^{(2)} \otimes e_2^{(2)} + e_1^{(2)} \otimes e_2^{(2)} \}$$ It isn't certain that you can rewrite any of these basis vectors as a simple tensor product.