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Inner-Product Spaces & Orthogonality

≈ 25 min
0/4 practice checks

An inner product ⟨u,v⟩\langle \mathbf{u}, \mathbf{v} \rangle generalises the dot product. In Rn\mathbb{R}^n it is ⟨u,v⟩=u1v1+⋯+unvn.\langle \mathbf{u}, \mathbf{v} \rangle = u_1 v_1 + \cdots + u_n v_n. From it we build:

  • Norm (length): ∥v∥=⟨v,v⟩.\|\mathbf{v}\| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle}.
  • Orthogonality: u⊥v\mathbf{u} \perp \mathbf{v} exactly when ⟨u,v⟩=0.\langle \mathbf{u}, \mathbf{v} \rangle = 0.
  • Projection of v\mathbf{v} onto u\mathbf{u}: proj⁡uv=⟨v,u⟩⟨u,u⟩ u.\operatorname{proj}_{\mathbf u}\mathbf{v} = \dfrac{\langle \mathbf{v}, \mathbf{u} \rangle}{\langle \mathbf{u}, \mathbf{u} \rangle}\, \mathbf{u}.

Worked example. For u=(1,2,2)\mathbf{u} = (1, 2, 2), ∥u∥=1+4+4=3.\|\mathbf{u}\| = \sqrt{1 + 4 + 4} = 3. And (1,1)⊥(1,−1)(1, 1) \perp (1, -1) because ⟨(1,1),(1,−1)⟩=1−1=0.\langle (1,1),(1,-1)\rangle = 1 - 1 = 0.

Two non-zero vectors are orthogonal exactly when their inner product equals…

Find the norm (length) ∥v∥\|\mathbf{v}\| of the vector v=(2,3,6)\mathbf{v} = (2, 3, 6).

For what value of kk are (1,2)(1, 2) and (4,k)(4, k) orthogonal?

When projecting v=(3,4)\mathbf{v} = (3, 4) onto u=(1,1)\mathbf{u} = (1, 1), the projection is c uc\,\mathbf{u}. Find the scalar coefficient c=⟨v,u⟩⟨u,u⟩.c = \dfrac{\langle \mathbf{v}, \mathbf{u} \rangle}{\langle \mathbf{u}, \mathbf{u} \rangle}.