Tensor products turn bilinear and multilinear constructions into ordinary linear maps. They are the bookkeeping device behind matrices, bilinear forms, exterior powers, and many coordinate-free constructions.
The universal property
Definition. A tensor product of vector spaces \(V\) and \(W\) is a vector space \(V\otimes W\) together with a bilinear map \(\tau:V\times W\to V\otimes W\), \((v,w)\mapsto v\otimes w\), such that for every bilinear map \(B:V\times W\to X\), there is a unique linear map \(\widetilde B:V\otimes W\to X\) with \[ B(v,w)=\widetilde B(v\otimes w). \]
Theorem. Tensor products exist and are unique up to unique isomorphism. If \((v_i)\) is a basis of \(V\) and \((w_j)\) is a basis of \(W\), then \((v_i\otimes w_j)_{i,j}\) is a basis of \(V\otimes W\). Hence \[ \dim(V\otimes W)=\dim V\dim W. \]
Proof. Construct \(V\otimes W\) with basis symbols \(v_i\otimes w_j\). Define \(v\otimes w\) by expanding \(v\) and \(w\) in bases and distributing coefficients. Any bilinear map is then determined by its values on pairs \((v_i,w_j)\), so it extends uniquely to a linear map from the constructed space. The universal property gives uniqueness.
Pure tensors and tensor rank
Definition. Elements of the form \(v\otimes w\) are pure tensors. General tensors are finite sums \(\sum_k v_k\otimes w_k\). The tensor rank is the smallest number of pure tensors needed in such a sum.
Counterexample. Not every tensor is pure. Under the identification \(F^m\otimes F^n\cong M_{m,n}(F)\), pure tensors correspond to rank-one matrices \(vw^\top\). A matrix of rank \(2\) is not a pure tensor.
Tensor products of maps
Proposition. If \(S:V\to V'\) and \(T:W\to W'\) are linear maps, there is a unique linear map \[ S\otimes T:V\otimes W\to V'\otimes W' \] satisfying \((S\otimes T)(v\otimes w)=S(v)\otimes T(w)\). In bases, its matrix is the Kronecker product.
Bilinear forms and tensors
Proposition. There are natural isomorphisms \[ \operatorname{Bilin}(V\times W,F)\cong (V\otimes W)^* \cong V^*\otimes W^* \] in finite dimension. Thus a bilinear form is the same thing as a linear functional on a tensor product.
Associativity and parentheses
Theorem. Tensor products are associative and distributive up to canonical isomorphism: \[ (U\otimes V)\otimes W\cong U\otimes(V\otimes W), \qquad U\otimes(V\oplus W)\cong (U\otimes V)\oplus(U\otimes W). \] These identifications are determined by the universal property.
Remark. The safest way to reason about tensor products is not by coordinates but by the universal property: to define a linear map out of \(V\otimes W\), define a bilinear map out of \(V\times W\).
Pure tensors in coordinates
Example. With standard bases of \(F^m\) and \(F^n\), the pure tensor \(x\otimes y\) may be represented by the rank-one matrix \(xy^\top\). A general tensor is a finite sum \[ \sum_{r=1}^k x_r\otimes y_r, \] which corresponds to a sum of rank-one matrices. The tensor product is not the set of pure tensors only; sums of pure tensors are essential.
Proposition. The bilinear map \((x,y)\mapsto xy^\top\) realizes \(F^m\otimes F^n\) as the vector space of \(m\times n\) matrices after bases have been chosen.
Proof. The matrices \(E_{ij}=e_i e_j^\top\) form a basis of the matrix space, and they are the images of \(e_i\otimes e_j\). Since the \(e_i\otimes e_j\) form a basis of \(F^m\otimes F^n\), the induced linear map is a bijection.