of A in U satisfies S 1 S 2 = { 0}, S 1 + S 2 = R n. Examples 1: (a.) S 1 = span . 1 (c.) S 1 = span, S , S 2 = span 0 (d.

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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 formulae 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 replace with + in the properties above, with U = R n. Definition.2. Two subspaces S, S 2 of R n are called complements if Examples : (a.) S = span {[ {[, S ]} ]} {[ {[ (b.) S = span, S ]} ]} S S 2 = { }, S + S 2 = R n. {[ ]} {[ ]} 7 (c.) S = span, S 8 (d.) S = span,, S 2 = span (e.) S = span,, S 2 = span,. Unlike for complements of finite sets, we did not define the{[ complement ]} of S {[. This ]} is because subsapces have infinitely many complements in general. For example, 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.. Let { v,..., v n } be a basis of R n, and let Then, S, S 2 are complements. 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 (See Section??, Problem??). Thus, S, S 2 are complements.

2 2 Morover, any two complements in R n are of this form, see Problem 6. Using in the special case of?? when S, S 2 are complements, reduces to dim(s + S 2 ) = dim(s ) + dim(s 2 ) dim(s S 2 ) dim(s ) + dim(s 2 ) = n, when S, S 2 are complements in R 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. (a) S + S 2 = R n (b) S S 2 = { } (c) dim(s ) + dim(s 2 ) = n. Proof. First, if (a), and (b) hold, then S, S 2 are complements by definition. If (a) and (c) hold, 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 (b) and (c) hold, 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. Theorem.. If S, S 2 are complements in R n, then each x R n can be written uniquely as x = v + w, for some v S, w S 2 Proof. We know x can be written in this way 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 this expression of x is unique. Definition.6. 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. The projection map π S,S 2 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. Also, by doing such a decomposition, we can find formula for projection maps. x x Examples 2: For each of the following, find formulae for π S,S 2 and π S,S 2. {[ {[ (a) S = span, S ]} ]} (b) S = span,, S 2 = span 2 2 x (c) S = span, S 2 = x + = Solutions:

3 (a) We write [ x ] = [ ] [ x + x2] = π S,S 2 ([ x ]) = [ ] ([ ]) [ x x, π S2,S =. x2] (b) We solve x = c + c 2 + c 2, which gives c = x /, = 2 /, = /. Plugging back in this solution, x x / / x x / x / = 2 / + 2 / = π S,S 2 = 2 /, π S2,S = 2 /. 2 (c) Since elements in S are scalar multiples of, we begin with x 2c x 2c = c + + c. c c x 2c In order for + c to be in S 2, we must have c (x 2c) + (2 + c) ( c) = = c = 6 ( x + ) Plugging this back in gives x ( x + ) = 2 ( x + x + ) + 6 ( x x + ) 6 x Therefore, π S,S 2 x ( x + ) = 2 (x + ), π S2,S 6 ( x x x + x = 2 x ) 6 x Notice that all of the above projection maps are linear maps. This is not a coincidence. We wouldn t be studying them in this class if they were not linear maps. Theorem.7. π S,S 2 : R n R n is a linear map with the following properties: (a) range(π S,S 2 ) = S, (b) ker(π S,S 2 ) = S 2, (c) π S,S 2 ( v) = v for all v S, (d) π 2 S,S 2 = π S,S 2. Proof. Let π = π S,S 2. First, for linearity, let 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, Also, for any scalar c, 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). π(c x) = π(c( v + w )) = π(c v + c w ) = c v = cπ( v + w ) = cπ( x). Thus, π is a linear map. For the kernel and range of π, we find range(π) = {π( v + w) v S, w S 2 } = { v v S } = S.

4 and 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. Next, for (c), all 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. Finally, for (d), we know that π( x) S for all x R n. Therefore, π(π( x)) = π( x) for all x R n = π π = π = π 2 = π. We have shown that if π is a projection map, then π 2 = π. However, the converse is true as well: If P : R n R n is a linear map satisfying P 2 = P, then P is a projection map. More specifically, we have the following. Proposition.8. If P : R n R n is a linear map which satisfies P 2 = P, then P is the projection onto range(p ) with respect to ker(p ). Proof. Problem 2. Exercises: () 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 = (g) S = x = x, = x, S 2 = x x + + = x (h) S = span 2, S 2 = x + x = x

5 (2) For the following complementary subspaces S, S 2, Find formulae for the projections π S,S 2 x x and π S2,S x. (a) S = { }, S 2 = R n {[ {[ 2 (b) S = span, S ]} 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]} ]} (b) S = ker(t ) = span 2, S 2 = span, {[ {[ (c) S = span, S 2]} 2]} (d) S = x x = =, S 2 = x x =

6 6 (e) S = x x =, S 2 = x x = = () Find all values of a so that S = span 2, S 2 = span 2, 6 a Problems: are complementary subspaces in R.. () Show that for complementary subspaces S, S 2 in R n, for all x R n. (π S,S 2 π S2,S ) ( x) = 2. The goal of this problem is to prove Theorem.8. 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 for all x R n. P ( x) = π range(p ),ker(p ) ( x).. (2) Suppose S, S 2 are complementary subspaces in R n, and T : R n R m is a linear map. Show that T π S,S 2 = T if and only if S 2 ker(t ).. (2) Suppose S, S 2 are complementary subspaces in R m, and T : R n R m is a linear map. Show that π S,S 2 T = T if and only if range(t ) S.. (2) Is it possible for to have complementary subspaces S, S 2 and S, S in R n so that If so, give an example. If not, why not? S S = S S = S 2 S = S 2 S = { }? 6. () 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 }. 7. () Suppose S, S 2 are complementary subspaces in R n, and V S is a subspace. Show that π (V ) = V + S 2. (See Section??, Problem?? for the definition of π (V )). 8. () Prove or give a counterexample: If P, P 2 : R n R n are projection maps, so is P P 2.