19.3. APPROXIMATIONS IN HILBERT SPACE 529

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.

This completes the proof.

Theorem 19.3.2 Let M be the span of {u1, · · · ,un} in a Hilbert space, H and let y ∈ H.Then Py is given by

Py =n

∑k=1

(y,uk)uk (19.3.17)

and the distance is given by √|y|2−

n

∑k=1|(y,uk)|2. (19.3.18)

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−

n

∑k=1

(y,uk)uk,u

)= 0

for all u ∈M and so by Corollary 19.1.13 this verifies 19.3.17.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 19.3.18.What if the subspace is the span of vectors which are not orthonormal? There is a

very interesting formula for the distance between a point of a Hilbert space and a finitedimensional subspace spanned by an arbitrary basis.