22.3. APPROXIMATIONS IN HILBERT SPACE 583

The vectors,{

u j}n

j=1 , generated in this way are therefore an orthonormal basis becauseeach vector has unit length.

Consider the second claim about finite dimensional subspaces. Without loss of gener-ality, assume {x1, · · · ,xn} is linearly independent. If it is not, delete vectors until a linearlyindependent set is obtained. Then by the first part,

span(x1, · · · ,xn) = span(u1, · · · ,un)≡M

where the ui are an orthonormal set of vectors. Suppose {yk} ⊆ M and yk → y ∈ H. Isy ∈M? Let yk ≡ ∑

nj=1 ck

ju j. Then let ck ≡(ck

1, · · · ,ckn)T

. Then

∣∣∣ck−cl∣∣∣2 ≡

n

∑j=1

∣∣∣ckj− cl

j

∣∣∣2 =( n

∑j=1

(ck

j− clj

)u j,

n

∑j=1

(ck

j− clj

)u j

)= ∥yk− yl∥2

which shows{ck}

is a Cauchy sequence in Fn and so it converges to c ∈ Fn. Thus

y = limk→∞

yk = limk→∞

n

∑j=1

ckju j =

n

∑j=1

c ju j ∈M. ■

Theorem 22.3.2 Let M be the span of {u1, · · · ,un} in a Hilbert space H and lety ∈ H. Then Py is given by

Py =n

∑k=1

(y,uk)uk (22.12)

and the distance is given by √|y|2−

n

∑k=1|(y,uk)|2. (22.13)

Proof: (y−

n

∑k=1

(y,uk)uk,up

)= (y,up)−

n

∑k=1

(y,uk)(uk,up)

= (y,up)− (y,up) = 0

It follows that (y−∑nk=1 (y,uk)uk,u) = 0 for all u ∈ M and so by Corollary 22.1.13 this

verifies 22.12.The square of the distance, d is given by

d2 =

(y−

n

∑k=1

(y,uk)uk,y−n

∑k=1

(y,uk)uk

)

= |y|2−2n

∑k=1|(y,uk)|2 +

n

∑k=1|(y,uk)|2

and this shows 22.13. ■