Linear Algebra Done Right · Problem 9D.12

Operators on Real Vector Spaces — Sheldon Axler

Mathematics — Linear Algebra · tensor-products

Problem

Suppose that V1,…,VmV_1, \ldots, V_m are finite-dimensional inner product spaces. Prove that there is a unique inner product on V1⊗⋯⊗VmV_1 \otimes \cdots \otimes V_m such that ⟨v1⊗⋯⊗vm,u1⊗⋯⊗um⟩=⟨v1,u1⟩⋯⟨vm,um⟩\langle v_1 \otimes \cdots \otimes v_m, u_1 \otimes \cdots \otimes u_m \rangle = \langle v_1, u_1 \rangle \cdots \langle v_m, u_m \rangle for all (v1,…,vm)(v_1, \ldots, v_m) and (u1,…,um)(u_1, \ldots, u_m) in V1×⋯×VmV_1 \times \cdots \times V_m. Note that the equation above implies that ∥v1⊗⋯⊗vm∥=∥v1∥×⋯×∥vm∥\|v_1 \otimes \cdots \otimes v_m\| = \|v_1\| \times \cdots \times \|v_m\| for all (v1,…,vm)∈V1×⋯×Vm(v_1, \ldots, v_m) \in V_1 \times \cdots \times V_m.
Intuitive explanation

The key is recognizing that an inner product on a tensor product must be defined first on simple tensors (products of individual vectors), then extended linearly—and the formula given is the only natural choice that respects the product structure. You'll need to verify that this extension is well-defined (independent of how you write a tensor as a sum of simple tensors), satisfies all inner product axioms, and is unique because simple tensors span the whole space.

Intuitive explanation

The key is recognizing that an inner product on a tensor product must be defined first on simple tensors (products of individual vectors), then extended linearly—and the formula given is the only natural choice that respects the product structure. You'll need to verify that this extension is well-defined (independent of how you write a tensor as a sum of simple tensors), satisfies all inner product axioms, and is unique because simple tensors span the whole space.

Related theory

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