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0 7M C= éc, ,czé’ fem m WM OIWOYmeaQ bags. +———— E. The Projection Theorem 1 Let V be an innerproduct space and let S be a subspace of V.
Fix v E V and consider the problem: inf llv r 3H (#) 565 In other words, we wish to ﬁnd an optimal approximation s in S to 1). It may happen that the infimum above is not achieved. In this case, we will say that
the optimization problem (#) does not admit a solution. It may happen that the infimum above is achieved. Any vector s that achieves the
inti mum will be called optimal (there may be many such vectors). Theorem (a) Suppose the optimization problem {#) is solvable. Then, s is optimal if and only
2'f
(v ~ s)J_S
Further, the optimal vector g is unique.
(b) Suppose S is complete. Then is solvable.
(c) Suppose S is ﬁnitedimensional. Let
B = {b1,b2,,bn}
be a basis for S and deﬁne the matrix M 6 cm“ by
<b1,b1> <b1,bn> M: . . ,
<bmb1> <bmbn> Then M is nonsingular. Moreover, is solvable and the unique solution is given. by n
S: E crib.
{:1 where
01 <b1,’U>
. = Ar‘ .
0,, <1)", u> —Rolla Proof: (a) Let é be optimal. We will show that (v — §)lS. Suppose not, i.e., there
exists 51 E S such that <sl,v—§>=a7é0 Deﬁne B = cr/llsﬂl2 and the vector snew = s + 551 E 5. Observe that [Iv—5mm“2 = llvSABSIH2
= v—§2+HB.91]2—— <v—§,le> — <le,v—§>
: llv—§H2+lﬁlzllslllzBa—ﬁa = “v  éllz lﬁ2512 which shows that s is not optimal. This contradicts our hypothesis, proving our claim
that (v — §)LS. To prove the reverse, let s be any vector that satisﬁes (1) — §)LS. We will show that
:3 be optimal. For this, examine llv—Sll2 = llU—é‘Jr§—6‘ll2
2+§—s2+ <v—§,§—s> + <§—s,'u—§> = llv — $1
The ﬁnal two inner products above are zero, since (1) — §)LS. This gives us
llv  3H2 = llv — éll2 + “3  5:“2 2 H12 — §2 for all s which establishes optimality of s. Finally, we will show uniqueness of the optimal vector. For this, let both be
optimal. Then, (we have just proved this) (1) — §)LS and (v — .§)LS. Now examine 2 —.e; 0:» l 
40:)
0:») I
Ch)
V 7—] which implies that — s = 0. This completes the proof. (b) Let S be complete. Deﬁne 'y 2 infses H1) — We have to establish that the
imfemum is achieved, or that there exists a vector 50 E S such that — so“ = 7. It is clear from the deﬁnition of 7 that for any 8 E 5, He — SH 2 7. Also, there exists a
sequence of vectors {8);} E S such that lim H1) — 8k“ 2 'y
k—ooo Now, using the parallelogram law, we get
“(Si — U) + (U — W“2 + “(32‘ * U) — (U  81)”2 = 2Si — UH2 + 2llsj — 11H2 This can be rearranged to give ”+%w ma—MF+Nn—UW*MW‘ 2 “Si — Sjllz S ZHSi — UHZ + lesj — 11H2 — 472
Then, using the fact that limk_.00 Ilv — 8);” = 'y, we get lim “3, — 3,“? g 272 + 272 — 472 : 0 k—ooo This forces limk_.00 “s, — sj = 0. Thus, the sequence of vectors {3k} 6 S is Cauchy.
Since 8 is complete, we are assured that this sequence converges to a vector 8" E 8. Finally, using continuity of norms, “11—80” = lim H'U—Skll =7 k—voo completing the proof. DD (c) We ﬁrst show that Al is nonsingular. Suppose not. Then, there exists a vector
0 79 a E C" such that Ma 2 0. This implies that n n
0:0‘N10222 a: <b¢,bj> aj = <q,q> = Iqu2
i=1j=l ' n
where q z aibi).
12:] As a consequence, q = Z ajbj = 0 (1)
1:1
Honrever, because [3 = {b1,b2, ~   ,b,,} is a basis, it is linearly independent. This contradicts ( l I, prOviug that M is invertible. Next, since 8 is ﬁnite dimensional, it is complete. Then, it is clear from part (b) of
the Projection Theorem that the optimization problem (#) is solvable. Since 8 is a
basis for S, we can write the optimal vector 5 E S as the linear combination n
5' = Z ajbj
jZI From part (a) the unique optimal vector s above must satisfy (v~—§)J_S, or equivalently,
<b,¢, (v — > = 0 for i = 1,  '  , n. Rewriting this gives 71 <b.,i,U> = <bj,.§> =2: <bi,bj> aj j=l
Stacking up these equations for i = 1,   ' , 71 yields
<b1,’U> <b1,‘U>
. 2 Ma or a = M‘1 _
<bn, v > <bn, v>
proving the claim. DD. 2 Example (Least Squares.)
We can employ the Projection theorem above to establish the following central result.
Theorem Let A E R’”"" have rank(/i) : n. Fin: 3; E R'" and consider the problem ﬁg}, “AI ‘ yll2 Then, the unique solution of this problem is
i = (A’A)_1A’y 3 Example (Fourier cosine series.) The classical Fourier series approximation may be viewed as an instance of the pro—
jection theorem. Consider the inner product space L2 [0, 2n]. Let
S : Span {1,cos(t),    ,cos(kt)} Given a function f(t) E L2[0,27r] we seek the optimal approximation in S of f.
Using the Projection theorem we obtain f(t) : a0 + a1 cos(t) +    + ak. c0s(lct) where 1 277 1 271’
‘10 _ “Wit, 01! = — f(t)cos(Zt)dt, Z = 1,2, . . . 7 k 27f 0 7T 0
These familiar formulae are the expressions for the Fourier cosine expansion of f. Deﬁnition Let 5 be a subspace of a Hilbert space V. The orthogonal complement
ofS is the set Si deﬁned by 5i={v€V: ILLS} Lemma {(1) Si is a subspace. (b) Si is complete. (C) (3i)i 2 5 (d) US is complete, then V = 8 EB Si (e) [f5 is complete, then (Si)i = 5
Proof: (a) Ly E 5i implies that < a:,q > = < y,q > = 0 for all q E S. Then,
<$+y,q>=0and <am,q>=0forallq€5. E] (b) Let 173° be a Cauchy sequence in Si. Since this is also a Cauchy sequence in V
which is a Hilbert space, we know that I], —> .r" E V. In order to show that Si is
complete, we have to show that the limit :5” above is in Si. For this, note that by continuity of the inner product, for any 3 E 5, <$°,s> = lim <mk,s> = lim 0:0
k—‘oo k—‘oo because 1']; E 5i. This proves that 1;” 6 Si. [:1 (e) Let s E S. Then, for all q E 5i, we have < q,s > = 0 which implies that
s E (SiH. D
(d) Fix 1) E V. Let s be the optimal approximation of v in 5. This exists by the
Projection Theorem on observing that 5 is complete. Then, we can write v=.§+(v—s) Note that from the Projection Theorem, (v — .§)_J_S or (v — E 5i. We have thus
shown that any vector in V can be expressed as the sum of a vector in 5 with one in SJ‘i All that remains is to show that this decomposition is unique. For this, suppose
we can write 11 E V as v=s+q251+q1 where 5,51 68 q,q1€SL
Then, 5 — $1 = q q1. But 8 and SJ are subspaces. Thus,
s—slzq—ql €808i
Then, we have
0: <s—51,q—q1> = <s—sl,s—sl> = “3—51”2 forcing s = .91 and q = q1, establishing uniqueness of the decomposition. D (e) All that remains is to show that (Si)i Q S. For this, let q E (SHi. Using the
directsum decomposition in (d)7 we can write q = 5 +7‘ with s E S and r 6 Si. Since
q E (SiH, we have that <q,7‘> = 0. Then, 0: <q,7‘> = <s,7‘> + <7‘,r> = <r,7‘> where the ﬁnal equality above follows because sir. It is then clear that 7' = O, forcing
q = s E S, proving the claim. (3 ...
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 Fall '10
 ClaireTomlin

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