x 2 For example, Theorem If S 1, S 2 are subspaces of R n, then S 1 S 2 is a subspace of R n. Proof. Problem 3.

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1 .. Intersections and Sums of Subspaces Until now, subspaces have been static objects that do not interact, but that is about to change. In this section, we discuss the operations of intersection and addition on subspaces. First, we define the intersection of two subspaces and show it is a subspace. We then describe how to intersect two subspaces using relation form and explore some alternative methods. Then, we define the sum of two subspaces and show how to sum two subspaces using span form. We end with an elegant relationship of how sum and intersection interact with dimension. Definition... The intersection of sets X and Y, denoted X Y, is the set of elements in X and Y : For example, X Y = {x x X, x Y }. {,, 5, 8, 9} {,, 7, 8} = {, 8}. Theorem... If S, S are subspaces of R n, then S S is a subspace of R n. Proof. Problem. Example... For each of the following pairs of subspaces S, S, find S S. ([ ]) ([ ]) (a) S = span, S = span, ([ ] [ ]) ([ ]) 5 (b) S = span,, S 7 = span, (c) S = span,, S = span,. 5 Solution: (a) We have so [ x ] S when = and [ x ] S when x =. Thus, { } S S =. We can also see the intersection S S in the following picture: S [ x ] S S when x = =, S S S ([ [ ]) 5 (b) Since, ] 7 Hence, S = R. As S R, is linearly independent, it spans R by the Invertible Matrix Theorem, Theorem.6.7. ([ S S = R S = S span. ])

2 Note the distinction between this intersection and {[ [ ]} 5 {[, =. ] 7 ]} (c) Unlike the first two examples, we cannot just draw a picture and see where S and S intersect. Geometrically, the intersection of these two non-identical planes in R is a line. But how do we figure out which line? It is not obvious how to find which nonzero vectors lie in span,, and span,. However, if we express S, S relation form, it is much easier to see what it means to 5 be in S S. Converting to relation form, S = x =, S = x x + = To have x x S, we need =, and to have S S, we need = AND x + =. In set notation, x S S = x + =, =. x S, we need x + =. Thus, to have More generally, to intersect two subspaces in relation form, just concatenate their relations! Then, by solving the linear system x + =, = to convert back to span form, we conclude S S = span. Concatenating the relations forms of two subspaces is the simplest way to intersect two subspaces, but it is not the only way. For the reader s potential convenience and to demonstrate the beauty of mathematics in having multiple ways to get the same answer, we present two other ways via to find intersections of subspaces by example, namely Example..(c). Depending on the circumstances, some of these methods will be more convenient than others, and we will use them occasionally in this textbook. The bottom line is you have some form or relation for each subspace, and you need to find when both of them hold. Method (Using one span form and relation form): We have x S = span,, S = x + = Suppose x S S. First, x S means c x = c + c = c for some scalars c, c. Now, x S means that the components satisfy x + =. Applying this c relation to x = c, we get c c = = c = c, c is free.

3 Thus, x must look like c x = c = c = S S = span. Method (Using span forms): We have S = span,, S = span, 5 Then, x S S when x = c + c = d + d 5 6 for some scalars c, c, d, d. Therefore, using multi-augmented matrix form, 5 6 In order to have a solution, we need d + d =, so let d = d and let d be free. Then, x = d + d 5 = d + d 5 = d d = d. 6 6 So, again, S S = span. Definition... The union of sets X and Y, denoted X Y, is the set of elements that lie in X or Y : X Y = {x x X or x Y }. For example, {,, 5, 8, 9} {,, 7, 8} = {,,, 5, 7, 8, 9}. How do intersection and union interact? We have () (X Y ) Z = (X Z) (Y Z), () (X Y ) Z = (X Z) (Y Z), () X Y = X + Y X Y. However, union is not an operation ([ ]) on subspaces, ([ because ]) the union of two subspaces need not be a subspace. For example, if S = span and S = span, then {[ ] } x S S = x x =, or =.

4 S [ ] [ / S ] S [ ] S [ ] [ ] [ ] Notice, S S, but / S S. Thus, S S is not closed under addition, so it is not a subspace. Instead of unioning, we will sum two subspaces. This sum operation on subspaces is the analog of union for finite sets in the sense that that sum S + S is the smallest subspace containing both S and S, Problem 7. Clearly, X Y is the smallest set containing both X and Y. Definition..5. If S, S are subspaces of R n, their sum S + S is the set of vectors that are some vector in S plus some vector in S : S + S = { x + y x S, y S }. Example..6. For each of the following pairs of subspaces S, S, find S + S. ([ ([ (a) S = span, S ]) = span, ]) (b) S = span,, S = span Solution: (a) For any [ x ] R, we have [ x ] = [ ] [ x + S x] + S, since [ ] [ x S and S x]. Hence, Note the distinction between S + S = R and S + S = R [ [ [ + =. ] ] ]

5 5 S S x + y S + S x S y S (b) Any vector x span, is of the form x = c + c for some scalars c, c. Any vector y span is of the form y = c for some scalar c. Thus, vectors in S + S are of the form x + y = c + c + c This means S + S = span,,. While intersecting subspaces is simpler in relation form, summing subspaces is simpler in span form. We hope Example..6(b) gives the reader a good idea for how to sum subspaces in general. As the reader might guess, we sum subspaces by concatenating their spanning sets. Proposition..7. If S, S are subspaces of R n, then S + S is a subspace of R n. Furthermore, span( u,..., u k ) + span( v,..., v m ) = span( u,..., u k, v,..., v m ).

6 6 Proof. If S, S are subspaces of R n, then by Theorem.., S, S can be written as the span of finite lists of vectors, say S = span( u,..., u k ) and S = span( v,..., v m ). Then, S + S = { x + y x S, y S } = {(c u + + c k u k ) + (d v + + d m v m ) c,... c k, d,... d m R} = {c u + + c k u k + d v + + d m v m c,... c k, d,... d m R} = span( u,..., u k, v,..., v m ). Thus, S + S is the span of a finite list of vectors, so S + S is a subspace as well by Theorem.. (a). Example..8. Consider the subspaces S, S, S R given by S = span, S = span, S = Find (S + S ) S and (S S ) + (S S ). w w w w =. Solution: Using Proposition..7 and converting to relation form, w S + S = span, = w w 6w w + w =, So, w (S + S ) S = w w 6w w w + w = w w w = w = w w 6w w + w =, w = w = w w w = 6 w, w =, w is free = 6 w w = w 6 /6 = span. On the other hand, notice c S S = c and Hence, w w w c w = = c c = = { }. w S S = c w c w w = = c c c = = { }. (S S ) + (S S ) = { } + { } = { }.

7 7 We have the operations of, and the notion of size for finite sets. We have the operations of, + and the notion of dimension for subspaces. By replacing by + and size by dimension, we can take the subspace analog of a property on finite sets to get what may or may not be a property on subspaces. For example, the subspace analog of (), (X Y ) Z = (X Z) (Y Z) for finite sets X, Y, Z, is () (S + S ) S? = (S S ) + (S S ) for subspaces S, S, S of R n. In this case, () is false, as Example..8 demonstrates. The subspace analog of () is also false, Problem 8. However, the subspace analog of () is true, Theorem..9. Theorem..9. For subspaces S, S of R n, Proof. Problem 6. dim(s + S ) = dim(s ) + dim(s ) dim(s S ), Exercises:. Compute the following intersections, expressing your answer in span form. {[ ] } {[ ] } x x (a) x x + = x x 5 = {[ ] } {[ ] } x x (b) = 5x x = ([ ]) ([ ]) (c) span span 6 x (d) x + + = (e) span, 5 x (f) span, span 5 x (g) span x 5 x x 5 + = x + =,. x + x = x (h) span 5, 7 x + x = x (i) span 5, 7 span, 5. x x (j) x x = + x = x x x x = x.

8 8. Compute the following sums, expressing your answer in span form. (a) span + span, 5. 7 ms x (b) x + x =, + x = + span 5 x. (c) x x =, = + x x + + =, x + 7 =. Problems:. () Show that if X Y and X Z, then X Y Z. () Show that if X Y, then X Z Y Z.. (+) Suppose S, S are subspaces of R n. Show that S S is also a subspace of R n by showing S S satisfies the three properties of a subspace.. Show that the sum and intersection operations on subspaces are commutative and associative. That is, show that for any subspaces S, S, S R n, (a) () S S = S S, (b) () (S S ) S = S (S S ), (c) () S + S = S + S, (d) () (S + S ) + S = S + (S + S ). 5. () Suppose S, S are subspaces of R n. Show that 6. Suppose S, S R n are subspaces with S S. (a) () Show that S S = S, (b) () Show that S + S = S. dim(s S ) dim(s ) + dim(s ) n 7. () Suppose that S, S are subspaces in R n, and S is a subspace of R n with S S and S S. Show that S + S S. 8. Find an example of subspaces S, S, S of R n for some n so that the subspace analog of () is false, i.e. (S S ) + S (S + S ) (S + S ). 9. () Suppose T, U : R n R m are linear maps. Prove or give a counterexample: range(t + U) = range(t ) + range(u).. Let T : R n R m be a linear map and S, S be subspaces of R n. (a) () Prove or give a counterexample: T (S + S ) = T (S ) + T (S ). (b) () Prove or give a counterexample: T (S S ) = T (S ) T (S ).. Let T : R n R m be a linear map and S, S be subspaces of R m. (a) () Prove or give a counterexample: T (S + S ) = T (S ) + T (S ). (b) () Prove or give a counterexample: T (S S ) = T (S ) T (S ).

9 9. () Suppose S, S,..., S k are subspaces of R n. Show that dim(s + S + + S k ) dim(s ) + dim(s ) + + dim(s k ).. () Suppose A is an m n matrix, and C is an m k matrix. Let [ A C ] be the m (n + k) matrix formed by attaching C directly to the right of A. Show that range ([ A C ]) = range(a) + range(c). [ A. () Suppose A is an m n matrix, and C is an k n matrix. Let be the (m + k) n matrix formed C] by attaching C directly below A. Show that ([ ]) A ker = ker(a) ker(c). C 5. Suppose S, S are subspaces of R n, ( u,..., u k ) is a basis for S, and ( v,..., v m ) is a basis for S. (a) () Show that S S = { } if and only if ( u,..., u k, v,..., v m ) is linearly independent. (b) () Use (a) to show that dim(s + S ) = dim(s ) + dim(s ) if and only if S S = { } without using Theorem The purpose of this problem is to prove Theorem..9. Suppose that S, S are subspaces of R n. Let ( u,... u k ) be a basis for S S. Extend it to a basis ( u,... u k, v,... v m ) for S, and let V = span( v,... v m ). Also, extend it to a basis ( u,... u k, w,... w r ) for S. (a) (+) Use Problem 5a to show that V (S S ) = { }. (b) (+) Show that V S = V (S S ). (c) () Show that S + S = V + S. (d) () What does Problem (b.) say about dim(v + S )? Now, verify that dim(s + S ) = dim(s ) + dim(s ) dim(s S ). 7. (a) () Show that for finite sets X, Y, Z, X Y Z = X + Y + Z X Y X Z Y Z + X Y Z. (b) () What is the subspace analog of (a)? Prove it or give a counterexample. 8. () Suppose S, S are subspaces of R n with S S = { }. Suppose ( u,..., u k ) is linearly independent in S, and w,..., w k S. Show that ( u + w,..., u k + w k ) is linearly independent. 9. (+) Suppose S, S are subspaces of R n. Show that S + S is a subspace by direct verification of the defining properties of being a subspace.. () Suppose S, S are subspaces of R n, and S S is also a subspace of R n. Show that S S or S S.. () Suppose S, S are subspaces of R n with S S = { }. Show that every v S + S can be written as for some unique x S and y S. v = x + y

10 . (+) Suppose T : R n R m is a linear map, and S = span( v,..., v k ) R m, and u,..., u k R n satisfy Show that T ( u ) = v,..., T ( u k ) = v k T (S) = span( u,..., u k ) + ker(t ).. () Suppose S,..., S k are subspaces of R n with bases B,..., B k, respectively. Show that the following are equivalent: (a) If v + + v k = with v S,..., v k S k, then v =,..., v k =. (b) Every x S + + S k can be written as x = v + + v k with v S,..., v k S k in a unique way. (c) B B k is linearly independent. (d) dim(s + + S k ) = dim(s ) + + dim(s k ). (e) S i (S + + S i + S i+ + + S k ) = { } for all i =,..., k.. () Suppose that A, A are both affine spaces. Recall that this means for all x, y A, and scalars t, we have t x + ( t) y A. The same holds for A. Show that A A is affine as well. 5. () Suppose that A, A are both affine spaces, and define Show that A + A is an affine space. A + A = { x + y x A, y A }