If $V,W$ say complex vector spaces, then it is well known fact there is a natural tensor product on $V \otimes W$ defined on the prime tensors by $$ \langle v \otimes w, v' \otimes w' \rangle = \langle v , v' \rangle \langle w ,w' \rangle.$$
The problem is to prove that this is acually an inner product.
Conjugation symmetry can be easily proven. However, I'm also need to prove, that it is positively definite. That is, for all $v_i \otimes w_i$ it holds that $$ \langle v_i \otimes w_i,v_i \otimes w_i\rangle \ge 0 $$ and $$ \langle v_i \otimes w_i,v_i \otimes w_i\rangle = 0 \iff v_i \otimes w_i $$ (here I use Einstein summation notation.) If one considers orthonormal bases $(e_i)_{i \in I}$ and $(f_j)_{j \in J}$ of $\mathrm{span}(v_i) \subset V$ and $\mathrm{span}(w_i) \subset W$ respectevely, then there are scalars $(\alpha)_{i\in I,j \in J}$ such that $v_i \otimes w_i = \alpha_{j,k}(e_j \otimes f_k)$. Then, it is clearly can be seen, that $$\langle v_i \otimes w_i,v_i \otimes w_i\rangle = |\alpha_{j,k}|^2 \ge 0. $$
However, I am interested if this fact can be proved without invoking any bases.
What I got by now is the identity
$$ \langle v_i \otimes w_i, v_i \otimes w_i \rangle = \sum^n_{i=1} \langle v_i, v_i \rangle \langle w_i, w_i \rangle + \sum^n_{j \neq i} \frac{1}{2}\left( \langle v_i, v_j \rangle \langle w_i, w_j \rangle + \overline{\langle v_i, v_j \rangle \langle w_i, w_j \rangle} \right) $$ which is clearly a real value, but I don't know how to prove that it is non-negative.