Solutions to Homework 11

Transcription

1 Solutions to Homework 11 Olena Bormashenko December 11, Check the 10 properties of vector spaces to see whether the following sets with the operations given are vector spaces. If they are vector spaces, give an argument for each property showing that it works; if not, provide an example (with numbers) showing a property that does not work. (a) V is the set of 2 2 matrices of the form [ ] 1 a A = 0 1 with the operations defined as A B = AB (normal matrix multiplication) c A = ca Note: Don t forget that these definitions only apply to the 2 2 matrices in the form I wrote above! This is NOT a vector space. It fails many of the properties, in particular, property (B) which states that c x must be in V for any x V. Here, letting [ ] 1 1 A = 0 1 and c = 2, we see that c A = ca = [ ] 2 2 / V 0 2 Note: I actually wanted this to be a vector space! I just goofed and defined scalar multiplication wrong... (b) V is R 2, with addition and scalar multiplication defined as follows: [ ] [ ] [ ] x1 y1 x1 + y = 1 x 2 y 2 x 2 + y 2 [ ] [ ] x1 2cx1 c = x 2 2cx 2 1

2 This is NOT a vector space. It fails property (7), which states that a (b u) = (ab) u. For example, if [ 1 u = 1] and a = b = 2, then we see that ( [ 1 a (b u) = 2 2 1]) [ 4 = 2 = 4] [ ] while [ [ 1 8 (ab) u = 4 = 1] 8] so we see that a (b u) (ab) u. (c) V is the set of functions from R to the positive real numbers: that is, V is the set of functions f with domain R such that f(x) > 0 for all x R. Let f and g be in V. We define f g and c f by defining the values these functions take on at any x R: (f g)(x) = f(x)g(x) (c f)(x) = f(x) c Note: In questions like this, it s good to be strict about your notation! Here, f and g are functions, while f(x) and g(x) are values that those functions take at x. So for example, notation like f(x) g(x) doesn t make any sense, because the operations and are only defined on the elements in the vector space, and not on the values the functions take. This IS a vector space. Let s check the properties. Since here, our vectors are functions, I ll be calling them f, g and h when checking the properties. The scalars will be c and d. (A) f g V : Clearly, f g is a function defined on all of R. Thus, the only thing to check is that (f g)(x) > 0 for all x. But we have that (f g)(x) = f(x)g(x) and since f(x) > 0 and g(x) > 0, we have that (f g)(x) > 0 for all x, as required. 2

3 (B) c f V : Since f(x) > 0, we have that (c f)(x) = f(x) c is defined for all x R, and furthermore, this is clearly positive. Therefore, c f is in V. (1) f g = g f: To check that two functions are equal, we need to check that they are the same at all x R. Thus, we need to check that (f g)(x) = (g f)(x) for all x R. We have that (f g)(x) = f(x)g(x) = g(x)f(x) = (g f)(x) so the property holds. (2) f (g h) = (f g) h: Again, we need to check that the functions are the same at all x R. If x R, then using the definition, (f (g h))(x) = f(x)(g h)(x) = f(x)g(x)h(x) = (f g)(x)h(x) = ((f g) h)(x) as required. (3) There exists an element 0 V such that 0 f = f = f 0 for all f: Here, we need to define the zero vector. Since this is a function, we need to state its value for all x R. Let z be the function such that z(x) = 1 for all x. Let us show that z is the zero vector for this vector space: for all x R, we have that (f z)(x) = f(x)z(x) = f(x) = (z f)(x) which shows that (f z) = f = (z f), as required. (4) There exists an element f V such that f ( f) = 0 = ( f) f for all f V : Since our 0 is the function z from above, we need to define f to fit it. We define ( f)(x) = 1 f(x) It is easy to check that this is a function in V, as its domain is R, and it only takes on positive values. Now, note that for x R, 1 (f ( f))(x) = f(x)( f)(x) = f(x) f(x) = 1 = z(x) = ( f)(x)f(x) = (( f) f)(x) Thus, for this definition of f, f ( f) = z = ( f) f, as required. (5) c (f g) = (c f) (c g): Let x R. Then, using the rules from above, (c (f g))(x) = (f g)(x) c = (f(x)g(x)) c = f(x) c g(x) c = (c f)(x)(c g)(x) = ((c f) (c g))(x) so the functions are equal for all x R, and the property holds. 3

4 (6) (c + d) f = (c f) (d f): Let x R. Then, ((c + d) f)(x) = f(x) c+d = f(x) c f(x) d = (c f)(x)(d f)(x) = ((c f) (d f))(x) so the functions are equal for all x R, and the property holds. (7) (cd) f = c (d f): Let x R. Then, ((cd) f)(x) = f(x) cd = (f(x) d ) c = [(d f)(x)] c as required. (8) 1 f = f: Let x R. Then, = (c (d f))(x) (1 f)(x) = f(x) 1 = f(x) so the property holds, as required. 2. Check whether the following sets W are non-empty subsets of V, closed under vector addition, and closed under scalar multiplication to check whether they are subspaces of the provided vector spaces V. If they are, show that everything works; if they are not, give an example that doesn t work. (You do NOT need to check whether V is a vector space!) Note: If V = R n, we assume that it has the standard vector addition and scalar multiplication, and similarly for V = M mn (the set of m n matrices.) Feel free to use the normal plus and scalar multiplication notation in those cases, instead of and. (a) Is W = {c 1 [1, 2] + c 2 [1, 1] c 1, c 2 R} a subspace of R 2? we re allowing vectors in R 2 to be row vectors.) (Here, This IS is a subspace. It s easy to check that it is a non-empty subset of R 2 (clearly, all the vectors in it have two coordinates, and for example, the vector [1, 2] is in W.) Thus, we need to check that it s closed under addition and closed under scalar multiplication. (1) x + y W for all x, y W : Let x and y be in W. Then, by definition of W, that means there exist scalars c 1, c 2, d 1, and d 2 such that Then, we see that x = c 1 [1, 2] + c 2 [1, 1] y = d 1 [1, 2] + d 2 [1, 1] x + y = (c 1 + d 1 )[1, 2] + (c 2 + d 2 )[1, 1] This is clearly in W since c 1 + d 1 and c 2 + d 2 are scalars. 4

5 (2) a x W for all a R, x W : Let x be in W. We can write it just like in part (1) above. Then, a x = a(c 1 [1, 2] + c 2 [1, 1]) = (ac 1 )[1, 2] + (ac 2 )[1, 1] which is clearly in W since ac 1 and ac 2 are scalars. (b) Is W = {the set of 2 2 diagonal matrices} a subspace of M 22? This IS is a subspace. It s clearly a subset of M 22, and it s clearly non-empty. Let us now check the other two subspace properties. Since the vectors in this vector space are matrices, we will use the letters A and B for them. (1) A + B W for all A, B W : Let A and B be in W. Then, we must have that [ ] [ ] a1 0 b1 0 A =, B = 0 a 2 0 b 2 for some a 1, a 2, b 1 and b 2. Then, we see that [ ] a1 + b A + B = a 2 + b 2 which is clearly a diagonal matrix, and hence is in W. (2) ca W for all c R, A W : Let A be in W, and let it be written like it is in part (1) above. Then, ca = c [ ] a1 0 = 0 a 2 [ ] ca1 0 0ca 2 which is clearly a diagonal matrix, and hence is in W. (c) Let W be the set of functions from [0, 1] to R such that f(0) = 1, and let V be the vector space of all functions from [0, 1] to R with the vector addition and scalar multiplication defined normally, that is: Is W a subspace of V? (f g)(x) = f(x) + g(x) (c f)(x) = cf(x) This is NOT a vector space. It s easy to check that the function f such that f(x) = x+1 is in W, while 2f is not in W, since 2f(0) = 2. Thus, property (2) doesn t hold. (Property (1) can be easily demonstrate not to hold, either.) 5

6 (d) Let A be an m n matrix, and let b be a vector in R m. For which values of { b is x A x = } b a subspace of R n? { Assume that W = x A x = } b is a subspace of R n. Then, if x 0 is in W, so is 2 x 0. Since x 0 is in W, we have that This means that A x 0 = b A(2 x 0 ) = 2A x 0 = 2 b But for 2 x 0 to be in W, we must by definition have A(2 x 0 ) = b. Therefore, we must have that b = 2 b which implies that our only possibility is b = 0. Now, let us show that if b = 0, then we indeed have a subspace. Let { } W = x A x = 0 Since A is m n, this is clearly a subset of R n, and since it contains 0 in R n, it s non-empty. Let us now check the other two properties of subspaces. (1) u + v W for all u, v W : Let u and v be in W. Then, by definition of W, we have that A u = 0 A v = 0 Therefore, using the properties of matrix-vector multiplication, A( u + v) = A u + A v = = 0 Since a vector is in W precisely if A times that vector is 0, this shows that u + v is in W, as required. (2) c u W for all c R, u W : Similarly to above, we have that Therefore, A u = 0 A(c u) = ca u = c 0 = 0 which shows that c u is in W, as required. 6

7 (e) Is W = {A M 22 A = 1} a subspace of M 22? This is NOT a subspace. For example, I 2 = [ ] satisfies I 2 = 1 and hence is in W, but the determinant of 2I 2 can be easily checked to be 4, and therefore, 2I 2 is not in W. This means that property (2) doesn t hold. 3. Prove the following statements using the properties of vector spaces at each step, name which property you re using to justify it. Be careful not to use intuitively obvious statements which aren t in the properties! (Yes, these are the statements that you were asked to show in class I m not sure most people got there, so I m giving them again.) (a) a 0 = 0 for all scalars a Proof: By property (3), = 0. Therefore, using property (5), a 0 = a( 0 + 0) = a 0 + a 0 By property (4), there exists an element ( a 0) (which we don t know to be the same as ( a) 0 it s merely notation for the additive inverse of a 0) such that a 0 + ( a 0) = 0. Thus, adding it to the right of the above equation, a 0 + ( a 0) = (a 0 + a 0) + ( a 0) Note that we have to put in the brackets to make sure that we re doing the exact same thing to both sides. Now, using property (2) to rearrange the right-hand side, and applying property (4), a 0 + ( a 0) = a 0 + (a 0 + ( a 0)) 0 = a Finally, by property (3), a = a 0, and hence as required. PHEW! 0 = a 0 7

8 (b) 0 v = 0 for all v V. Proof: This proof is pretty similar to the last one. Clearly, since 0 is just a scalar, = 0. Therefore, using property (6), 0 v = (0 + 0) v = 0 v + 0 v By property (4), there exists a vector 0 v such that 0 v + ( 0 v) = 0. Adding it to both sides on the right, 0 v + ( 0 v) = (0 v + 0 v) + ( 0 v) Rearranging using property (2), and simplifying the left-hand side with property (4), 0 = 0 v + (0 v + ( 0 v)) Finally, using property (4), then property (3), as required. (c) ( 1) v = v for all v V. 0 = 0 v + 0 = 0 v Proof: By property (8), 1 v = v. Therefore, v + ( 1) v = 1 v + ( 1) v Using property (6), 1 v + ( 1) v = (1 + ( 1)) v = 0 v. From part (b) above, we know that 0 v = 0. Thus, v + ( 1) v = 0 From property (4), there exists a vector v (not necessarily equal to ( 1) v that s what we have to prove!) such that ( v) + v = 0. Adding it to both sides on the left, we get ( v) + ( v + ( 1) v) = ( v) + 0 Using property (2) on the left, and property (3) on the right, Using property (4), and finally, using property (3), as required. (( v) + v) + ( 1) v = v 0 + ( 1) v = v ( 1) v = v 8