$𝕂$ -tensor product of vector spaces

Let $𝑉 1, 𝑉 2$ be vector spaces over $𝕂$ . The tensor product $𝑉 1 \otimes 𝕂 𝑉 2$ is a vector space which allows one to treat $𝕂$ -bilinear maps from $𝑉 1 \times 𝑉 2$ as $𝕂$ -linear maps from $𝑉 1 \otimes 𝕂 𝑉 2$ , as ensured by the Universal property. This is a special case of the $K$ -tensor product of modules, whose universal property should be taken as the definition.

Constructions

Here we give alternate constructions which may be more convenient than the construction as a quotient of a free module.

Finite dimensional vector spaces

The tensor product $𝑉 \otimes 𝑊$ of finite-dimensional vector spaces $𝑉, 𝑊$ over a field $𝕂$ is the vector space of bilinear forms $𝑉 * \times 𝑊 * \to 𝕂$ , equipped with a bilinear¹ map $(\otimes) : 𝑉 \times 𝑊 \to 𝑉 \otimes 𝑊$ such that #m/def/linalg

(𝑣 \otimes 𝑤) (𝑓, 𝑔) = 𝑓 (𝑣) 𝑔 (𝑤)

for $𝑣 \in 𝑉$ , $𝑤 \in 𝑊$ , $𝑓 \in 𝑉 *$ , $𝑔 \in 𝑊 *$ .²³ It follows that if ${𝑣 𝑖} 𝑛 𝑖 = 1$ and ${𝑤 𝑗} 𝑚 𝑗 = 1$ are bases of $𝑉$ and $𝑊$ respectively, then ${𝑣 𝑖 \otimes 𝑤 𝑗} 𝑛, 𝑚 𝑖, 𝑗 = 1$ defines a basis for the tensor product space $𝑉 \otimes 𝑊$ . We call

𝑇 \in 𝑉 \otimes \dots \otimes 𝑉 ⏟__⏟__⏟ 𝑝 \otimes 𝑉 * \otimes \dots \otimes 𝑉 * ⏟__⏟__⏟ 𝑞 = T 𝑝 𝑞

a type $(𝑝, 𝑞)$ Tensor⁴

Additional structure

Hilbert spaces

If $𝑉, 𝑊$ are finite-dimensional Hilbert spaces. then the tensor product $𝑉 \otimes 𝑊$ is a Hilbert space carrying the unique inner product given by

⟨ 𝑣 \otimes 𝑤 | 𝐵 ⟩ = 𝐵 (⟨ 𝑣 |, ⟨ 𝑤 |)

Then if ${𝑣 𝑖} 𝑛 𝑖 = 1$ and ${𝑤 𝑗} 𝑚 𝑗 = 1$ are orthonormal bases of $𝑉, 𝑊$ respectively, ${𝑣 𝑖 \otimes 𝑤 𝑗} 𝑛, 𝑚 𝑖, 𝑗 = 1$ is an orthonormal basis of $𝑉 \otimes 𝑊$ .

Properties

$d i m (𝑉 \otimes 𝑊) = d i m 𝑉 \cdot d i m 𝑊$

𝕂-tensor product of vector spaces