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Exercise 246Linear algebra › Inner product spacesHigh schoolAdded 3 Nov 2026

Inner product axioms

#inner product#axioms#dot product

Statement

Let E=R2E=\mathbb{R}^2 with the map ⟨u,v⟩=2u1v1+u2v2\langle u,v\rangle = 2u_1v_1 + u_2v_2. Show that ⟨⋅,⋅⟩\langle\cdot,\cdot\rangle is an inner product on R2\mathbb{R}^2.

Hint

Check the four axioms: bilinearity, symmetry, and positive definiteness.

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Solution

Bilinearity. For u,v,w∈R2u,v,w\in\mathbb{R}^2 and λ∈R\lambda\in\mathbb{R}:
⟨u+λw,v⟩=2(u1+λw1)v1+(u2+λw2)v2=⟨u,v⟩+λ⟨w,v⟩.\langle u+\lambda w,v\rangle = 2(u_1+\lambda w_1)v_1+(u_2+\lambda w_2)v_2 = \langle u,v\rangle+\lambda\langle w,v\rangle.
Symmetry in the second variable follows similarly.

Symmetry. ⟨u,v⟩=2u1v1+u2v2=2v1u1+v2u2=⟨v,u⟩\langle u,v\rangle = 2u_1v_1+u_2v_2 = 2v_1u_1+v_2u_2=\langle v,u\rangle.

Positive definiteness. ⟨u,u⟩=2u12+u22≥0\langle u,u\rangle = 2u_1^2+u_2^2\ge0, with equality iff u1=u2=0u_1=u_2=0.

Conclusion. ⟨⋅,⋅⟩\langle\cdot,\cdot\rangle is an inner product on R2\mathbb{R}^2.