112 CHAPTER 4. LINEAR SPACES

Also, since (u− z,v) = 0 for all v ∈V, |u|2 = |u− z+ z|2 = |u− z|2 + |z|2 so

|u|2 =

∣∣∣∣∣u− n

∑i=1

zivi

∣∣∣∣∣2

+

∣∣∣∣∣ n

∑i=1

zivi

∣∣∣∣∣2

= d2 +

∣∣∣∣∣ n

∑i=1

zivi

∣∣∣∣∣2

= d2 +∑j

=(u,v j)︷ ︸︸ ︷∑

i(v j,vi)ziz j = d2 +∑

j(u,v j)z j

= d2 +yTz, y ≡ ((u,v1) , · · · ,(u,vn))T , z ≡

(z1, · · · ,zn)T

From ∗, Gz = y,

(G(v1, ...,vn) 0

yT 1

)(zd2

)=

(y

∥u∥2

). Now use Cramer’s

rule to solve for d2 and get

d2 =

det(

G(v1, ...,vn) y

yT |u|2)

det(G(v1, ...,vn))≡ detG(v1, ...,vn,u)

detG(v1, ...,vn)

23. In the situation of Problem 21, let fk (x)≡ xk and let V ≡ span( fp1 , ..., fpn). give anestimate for the distance d between fm and V for m a nonnegative integer and as inthe above problem − 1

2 < p1 < · · · < pn. Use Theorem 1.9.28 in the appendix andthe above problem with vi ≡ fpi and vn+1 ≡ fm. Justify the following manipulations.

The numerator in the above formula for the distance is of the form ∏ j<i≤n+1(pi−p j)2

∏i, j≤n+1(pi+p j+1)

=∏ j<i≤n (pi− p j)

2∏ j≤n (m− p j)

2

∏i, j≤n (pi + p j +1)∏ni=1 (pi +m+1)∏

nj=1 (p j +m+1)(2m+1)

While G( fp1 , ..., fpn) =∏ j<i≤n(pi−p j)

2

∏i, j≤n(pi+p j+1). Thus d =

∏ j≤n|m−p j|∏

ni=1(pi+m+1)(

√2m+1)

.

24. Suppose ∑nk=0 aktk = 0 for each t ∈ (−δ ,δ ) where ak ∈ X , a linear space. Show that

each ak = 0.

25. Suppose A ⊆ Rp is covered by a finite collection of Balls F . Show that then thereexists a disjoint collection of these balls, {Bi}m

i=1, such that A⊆∪mi=1B̂i where B̂i has

the same center as Bi but 3 times the radius. Hint: Since the collection of balls isfinite, they can be arranged in order of decreasing radius. Mimic the argument forVitali covering theorem.