Notice that the set complement of A in U satisfies

Transcription

1 Complements and Projection Maps In this section, we explore the notion of subspaces being complements Then, the unique decomposition of vectors in R n into two pieces associated to complements lets us define projection maps Finally, we explore how to find formulas for projection maps and study their properties In order to find the right definition of subspace complements, we refer back to the definition of complements of sets Definition Given a set U, the set complement of A in U, denoted A C is given the set of all elements that are in U but not in A: A C = {x U x / A} For example, if U = {, 2,,,, 6} and A = {2,, 6}, then A C = {,, } () Notice that the set complement of A in U satisfies A C A =, A C A = U Here, the set complement of a subspace in R n is not a subspace Instead to find the right notion of a subspace complement, we set U = R n and replace with + in () Definition 2 Two subspaces S, S 2 of R n are called complements (in R n ) if S S 2 = { }, S + S 2 = R n Example The following pairs of subspaces are complements ([ ([ (a) S = span, S ]) 2 = span ]) (b) S = span, S 2 = span 7 (c) S = span, S 2 = span 8 (d) S = span,, S 2 = span (e) S = span,, S 2 = span, Solution: We leave it to reader to verify that these pairs of subspaces are complementary, Exercise Unlike for complements of finite sets, we did NOT define the complement of S This is because ([ subspaces ]) have infinitely many complements in general For example, from Example (a), (b), span and ([ ([ ([ span are both complements of span In fact, every -dimensional subspace except span ]) ]) ]) ([ is a complement of span We now demonstrate a very general way to construct complementary ]) subspaces Proposition Suppose ( v,, v n ) is a basis of R n Then, are complements in R n Proof First, S = span( v,, v k ), S 2 = span( v k+,, v n ) S + S 2 = span( v,, v k ) + span( v k+,, v n ) = span( v,, v n ) = R n Also, S S 2 = because { v,, v k, v k+,, v n } is linearly independent, using Section??, Problem 2a) Thus, S, S 2 are complements

2 2 Moreover, any two complements in R n are of this form, Problem 2 Using in the special case of Theorem 9 when S, S 2 are complements in R n, reduces to dim(s + S 2 ) = dim(s ) + dim(s 2 ) dim(s S 2 ) dim(s ) + dim(s 2 ) = n, In fact, the property dim(s ) + dim(s 2 ) = n can replace either of the defining properties of complements, but not both Proposition Any two of the following make S, S 2 complements in R n (i) S + S 2 = R n (ii) S S 2 = { } (iii) dim(s ) + dim(s 2 ) = n Proof First, if (i) and (ii) are true, then S, S 2 are complements by definition If (i) and (iii) are true, then dim(s S 2 ) = dim(s ) + dim(s 2 ) dim(s + S 2 ) = n n = = S S 2 = { }, so S, S 2 are complements If (ii) and (iii) are true, then dim(s + S 2 ) = dim(s ) + dim(s 2 ) dim(s S 2 ) = n = n = S + S 2 = R n so S, S 2 are complements One of the most important properties of complementary subspaces in R n is the following decomposition Theorem, Theorem 6 Breaking up complicated objects into simpler pieces is common in mathematics and other areas, and Theorem 6 is an example of this Theorem 6 also allows us to define projection maps, which show up often in mathematics and other areas Theorem 6 If S, S 2 are complements in R n, then each x R n can be written uniquely as Proof We know x = v + w, for some v S, w S 2 (2) x = v + w, for some v S, w S 2 since x R n = S + S 2 For uniqueness, suppose that x = v + w = v 2 + w 2 for some v, v 2 S, w, w 2 S 2 Then we have v v 2 = w 2 w S S 2 = { } since v v 2 S and w 2 w S 2 This means v = v 2 and w = w 2, so v, w are unique in (2) Definition 7 Given complementary subspaces S, S 2 of R n, we define the projection map π S,S 2 : R n R n onto S with respect to S 2 as follows: π S,S 2 ( v + w) = v for all v S, w S 2 Definition 7 makes sense since each x R n can be written uniquely as x = v+ w for some v S, w S 2, so we are liberty to use this form for the input In addition, if x = v + w for some v S, w S 2, then we have This means for all x R n, π S,S 2 ( x) = v, π S2,S ( x) = w x = π S,S 2 ( x) + π S2,S ( x)

3 S 2 v w = π S2,S ( x) S 2 x w v = π S,S 2 ( x) S S Thus, by writing x = v + w where v S, w S 2, we can find formulae for π S,S 2 ( x) and π S2,S ( x) Example 8 For each of the following, find formulae for π S,S 2 (a) S = span (b) S = span (c) S = span ([, S 2 = span ]), 2, S 2 = span 2, S 2 = x x + = x and π S2,S x Solutions: (a) We write (b) We solve () which gives [ x ] = [ ] [ x + x2] = π S,S 2 ([ x ]) = x = c + c 2 + c 2, c = x /, c 2 = 2 /, c = / [ ] [ x x, π S2,S = x2] Plugging this solution back into (), x x / / = 2 / + 2 / = π S,S 2 x / = 2 /, π S2,S x / = 2 / x

4 (c) Since elements in S are scalar multiples of 2, we begin with x = 2c c + x 2c () + c c c 2c x 2c where c S, and we want + c S 2 In order to have c c x 2c + c S 2, we need c (x 2c) + (2 + c) ( c) = = c = 6 ( x + ) Plugging this back into (), x ( x + ) = 2 ( x + x + ) + 6 ( x x + ) 6 x Therefore, π S,S 2 x ( x + ) x = 2 ( x + x + ), π S2,S 6 ( x = 2 x ) 6 x The projection maps in Example 8 are linear maps This is not a coincidence We wouldn t be studying them in this class if they were not linear maps In addition, projection maps important properties that are worth specifying here In particular, the range and kernel of π S,S 2 are easy to describe in terms of S, S 2 Doing a projection map twice is the same as doing it once Finally, we realize that this last property characterizes projection maps! Proposition 9 Suppose S, S 2 R n are complementary subspaces Then, π S,S 2 following properties: (a) π S,S 2 is linear : R n R n has the (b) range(π S,S 2 ) = S, (c) ker(π S,S 2 ) = S 2, (d) π S,S 2 ( v) = v for all v S, (e) π 2 S,S 2 = π S,S 2 Proof For brevity, let π = π S,S 2 (a) For linearity, suppose x, y R n and write x = v + w, y = v 2 + w 2 for some v, v 2 S and w, w 2 S 2 Then, because S, S 2 are subspaces, v + v 2 S and w + w 2 S 2, so π( x + y) = π(( v + w ) + ( v 2 + w 2 )) = π(( v + v 2 ) + ( w + w 2 )) = v + v 2 = π( v + w ) + π( v 2 + w 2 ) = π( x) + π( y) Fix a scalar c Because S, S 2 are subspaces, c v S and c w S 2, so π(c x) = π(c( v + w )) = π(c v + c w ) = c v = cπ( v + w ) = cπ( x) (b) Thus, π is linear range(π) = {π( v + w) v S, w S 2 } = { v v S } = S

5 (c) ker(π) = { v + w v S, w S 2, π( v + w) = } = { v + w v S, w S 2, v = } = { w w S 2 } = S 2 (d) Any v S can be written as v = v +, where S 2, so using the definition of projection maps, π( v) = π( v + ) = v for all v S (e) By definition, π( x) S for all x R n Therefore, using Proposition 9(c), π(π( x)) = π( x) for all x R n = π π = π = π 2 = π Theorem If P : R n R n is linear map and satisfies P 2 = P, then P is the projection onto range(p ) with respect to ker(p ) Proof Problem 7 Exercises: Verify that the pairs of subspaces in are complementary 2 Determine if the following pairs of subspaces are complementary Explain ([ (a) S = span, S ]) 2 = span ([ ] [ ]) (b) S = span, S 2 = span, (c) S = span, S 2 = span (d) S = span, S 2 = span, 2 (e) S = span 2, S 2 = span 7, x (f) S = span 2, S 2 = x = x x (g) S = = x, = x, S 2 = x + + = x (h) S = span 2 2 = x + x = x

6 6 For the following complementary subspaces S, S 2, Find formulas for the projections π S,S 2 x and x π S2,S x (a) S = { }, S 2 = R n ([ 2 (b) S = span, S 2 = span 7]) (c) S = span, S 2 = w w 2 w w 2w 2 + w = (d) S = span 2, S 2 = span, 2 2 Find complements of the following subspaces (a) span 2, (b) span 2, 6 (c) span 2 (d) span, (e) span 2 For each of the following subspaces S, S 2, Find a linear map T : R n R n with the appropriate n so that ker(t ) = S and range(t ) = S 2 ([ ([ (a) S = span, S 2]) 2 = span ]) (b) S = ker(t ) = span 2 ([ (c) S = span, S 2]) 2 = span, S 2 = span ([ 2]), (d) S = x x = =, S 2 = x x = x x (e) S = x =, S 2 = x = =

7 7 6 Find all values of a so that 2 S = span 2, S 2 = span, 6 a are complementary subspaces in R Problems: (+) Suppose S R n is a subspace Show that there exists a subspace S 2 R n so that S, S 2 are complementary 2 (+) Suppose S, S 2 R n are complementary subspaces Show that id n = π S,S 2 + π S2,S (2) Suppose S, S 2 R n are subspaces with S S 2 = { } Show that there exists a subspace S of R n so that S 2 S and S, S are complementary () Suppose S, S 2 R n are subspaces with S + S 2 = R n Show that there exists a subspace S of R n so that S S 2 and S, S are complementary (2) Suppose S, S 2 R n are complementary subspaces Show that π S,S 2 π S2,S = O nn 6 (2) Suppose S, S 2 R n are subspaces so that every x R n can be expressed as Show that S, S 2 are complementary x = v + w, for some unique v S, w S 2, 7 The goal of this problem is to prove Theorem Suppose that P : R n R n is a linear map which satisfies P 2 = P (a) () Show that ker(p ) and range(p ) are complements in R n (b) () Show that P = π range(p ),ker(p ) 8 () Suppose S, S 2 R n are complementary subspaces, and T : R n R m is linear Show that T π S,S 2 = T if and only if S 2 ker(t ) 9 () Suppose S, S 2 R n are complementary subspaces, and T : R n R m is linear Show that π S,S 2 T = T if and only if range(t ) S (2) Is it possible for to have subspaces S, S 2, S, S of R n so that S, S 2 are complementary, S, S are complementary, and S i S j = { } for all i j? If so, give an example If not, why not? () Suppose S, S 2, S are subspaces of R n with S, S 2 complementary Show that π S2,S (S ) = S 2 if and only if S + S = R n 2 () Suppose S, S 2 are complements in R n Show that there exists a basis { v,, v n } of R n so that S = span( v,, v k ), S 2 = span( v k+,, v n ) () Suppose S, S, S 2 R n are subspaces, with S, S 2 complementary Show that π S,S 2 (S) = (S + S 2 ) S

8 8 () Suppose S, S, S 2 R n are subspaces, with S, S 2 complementary Show that π S,S 2 (S) = (S S ) + S 2 (2) Prove or give a counterexample: If P, P 2 : R n R n are projection maps, so is P + P 2 6 (2+) Prove or give a counterexample: If P, P 2 : R n R n are projection maps, so is P P 2 7 Suppose S, S 2 R n are complementary subspaces, and T : R n R m is linear Let U = T π S,S 2 Show that (a) (+) range(u) = T (S ) (b) (2) ker(u) = (ker(t ) S ) + S 2 8 Suppose S, S 2 R n are complementary subspaces, and T : R n R m is linear Let U = π S,S 2 T Show that (a) (2) ker(u) = T (S 2 ) (b) (2) range(u) = (range(t ) + S 2 ) S 9 Suppose S,, S k R n are subspaces so that every x R n can be expressed as x = v + + v k, for some unique v S,, v k S k (a) (2) Show that for each j =, k, S j and S + + S j + S j+ + + S k are complementary subspaces in R n (b) (2+) For each j =, k, let π j = π Sj,S + +S j +S j++ +S k Show that x = π ( x) + + π k ( x) 2 Suppose S, S 2, S, S R n are subspaces so that S, S 2 are complements, S, S are complements, S S, and S S 2 (a) () Show that if y S 2, then π S,S ( y) S 2 (b) () Show that for all x R n, there exists unique u S, v S 2 S, w S so that (c) (2) Show that π S,S 2 π S,S = π S,S 2 x = u + v + w