Page 126 - Linear Algebra Done Right

Suppose U is a subspace of V. The decomposition V = U ⊕U given
by 6.29 means that each vector v ∈ V can be written uniquely in the
form Orthogonal Projections and Minimization Problems ⊥ 113
v = u + w,
where u ∈ U and w ∈ U . We use this decomposition to define an op-
⊥
erator on V, denoted P U , called the orthogonal projection of V onto U.
For v ∈ V, we define P U v to be the vector u in the decomposition above.
In the notation introduced in the last chapter, we have P U = P U,U ⊥. You
should verify that P U ∈L(V) and that it has the following proper-
ties:
• range P U = U;
• null P U = U ;
⊥
• v − P U v ∈ U ⊥ for every v ∈ V;
2
• P U = P U ;
•HP U v ≤ v for every v ∈ V.
Furthermore, from the decomposition 6.31 used in the proof of 6.29
we see that if (e 1 ,...,e m ) is an orthonormal basis of U, then
6.35 P U v = v, e 1 e 1 +· · ·+Pv, e m e m

for every v ∈ V.
The following problem often arises: given a subspace U of V and
a point v ∈ V, find a point u ∈ U such that v − u is as small as
possible. The next proposition shows that this minimization problem
is solved by taking u = P U v.
6.36 Proposition: Suppose U is a subspace of V and v ∈ V. Then The remarkable
simplicity of the
v − P U v ≤ v − u
solution to this
for every u ∈ U. Furthermore, if u ∈ U and the inequality above is an minimization problem
equality, then u = P U v. has led to many
applications of
Proof: Suppose u ∈ U. Then inner-product spaces
outside of pure
2
2
2
6.37 v − P U v ≤ v − P U v + P U v − u
mathematics.
6.38 = (v − P U v) + (P U v − u) 2
2
= v − u ,