Lecture 31 - Last Lecture!

We'll do some cool things with vector projections today!

Consider:

We'll define vector projection in an abstract sense. Recall for our purposes:

Lemma

Suppose U is a finite-dimensional subspace of V. Then:
V=U⊕U⊥
as well as:
(U⊥)⊥=U

We now show a cool definition:

orthogonal projection

Suppose U is a finite-dimensional subspace of V, the orthogonal projection of V onto U is the operator Pu, defined by:

PU(v→)=u

whenever v=u+w where u∈U and w∈U⊥.

Properties of orthogonal projection

  • PU∈L(V);
  • PU(u)=u;
  • PU(w)=0 for w∈U⊥;
  • range(PU)=U;
  • null(PU)=U⊥;
  • v−PU(v)∈U⊥;
  • (PU)2=PU;
  • ‖PU(v)‖≤‖v‖;
  • Given any orthonormal basis for the space U, say e1,...,em, then PU(v)=∑i=1m⟨v,ei⟩ei (this is the v−PU(v)) idea coming from Gram-Schmidt.

We'll prove some of these. Let's prove ‖PU(v)‖≤‖v‖:

Proof
Suppose v=u+w where u∈U and w∈U⊥. Then:

‖PU(v)‖2=‖u‖2≤‖u‖2+‖w‖2=‖u+w‖2=‖v‖

Thus taking both sides reveals the theorem.
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Minimization Problems

We sometimes want to find the minimum norm between two vectors. For some setup, suppose U is a finite-dimensional subspace of V. Then:

‖v−PU(v)‖≤‖v−u‖

for all u∈U. Namely, the distance from the "plane" U, is less than just taking the distance fr

Proof

‖v−PU(v)‖2≤‖v−PU(v)‖2⏟∈U⊥+‖PU(v)−u‖2⏟∈U=‖v−u‖2

Which comes from using the Pythagorean Theorem along with the fact that the two vectors are orthogonal.
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So the best vector to use as an approximation of v, using only vectors in U, would be Pu(v).

An Example

See HW 7 - Inner Product Spaces#^af860a problem. The list is orthonormal with respect to:

⟨f,g⟩=∫−ππf(x)g(x)dx

Let Un=span(Bn), where Bn was up to the cos⁡(nx) and sin⁡(nx) terms from our list. Un is a finite-dimensional subspace of all continuous real valued functions on −π to π.

Note that then we can compute PU1(ex):

U1=span(12π,cos⁡(x)π,sin⁡(x)π)

So then:

PU1(ex)=⟨ex,12π⟩12π+…