# Inner product of a tensor product

Say we have a space $A$ and $B$ with a basis $|a_i>$ and $|b_i>$ on it. On the space $A \otimes B$, $|a_i> \otimes |b_j>$ is a basis vector. I am having difficulty understanding what happens when we take an inner product on this space. For example, $$(<a_i| \otimes <b_j| )(|a_i> \otimes |b_j>) \overset{?}{=} <a_i|a_i> \otimes <b_j |b_j>$$

What does a tensor product of two numbers mean? Or more general the tensor product of a number with an vector in one of the spaces? This appears in the context of the partial trace.

• See Wikipedia. Dec 31, 2017 at 14:04
• Wikipedia is enough. However, from the universality property of the tensor product you see that $c\otimes {\bf v} = c{\bf v}$ and $c\otimes c' = c\cdot c'$ (where the identities are natural isomorphisms of vector spaces)... This sort of questions are however more appropriate for mathSE... Dec 31, 2017 at 15:00
• Universality says that the bi-linear map $f: {\mathbb R} \times {\mathbb R}\ni (c,c') \mapsto c\cdot c'\in \mathbb R$ uniquely "extends" (in the sense of univerality theorem) to a linear map $F: {\mathbb R}\otimes {\mathbb R} \mapsto \mathbb R$. This map is evidently surjective because its restriction $f$ is already surjective. Finally notice that ${\mathbb R}\otimes {\mathbb R}$ has dimension $1 \times 1 =1$ as a real vector space and $\mathbb R$ as dimension $1$ as well. Dec 31, 2017 at 16:35
• A linear surjective map from a finite dimensional vector space onto a finite dimensional vector space with the same dimension is necessarily injective. Therefore $F$ is an isomorphism. Dec 31, 2017 at 16:35
• As a matter of fact, the universality theorem applied to this case says nothing but that the isomorphism $F: {\mathbb R}\otimes {\mathbb R} \mapsto \mathbb R$ is the unique linear extension of $c \otimes c' \mapsto c\cdot c'$ for every $c,c' \in \mathbb R$. Dec 31, 2017 at 16:38