Math 110: Worksheet 1 Solutions

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1 Math 110: Worksheet 1 Solutions August 30 Thursday Aug Determine whether or not the following sets form vector spaces over the given fields. (a) The set V of all matrices of the form where a, b R, over R with standard addition and scalar multiplication. and Note that V is not closed under addition: for a, b, c, d R, we have 1 c but + 1 c ( ) a + c b + d We conclude that V is not a vector space with the given operations. (b) The set V of all matrices of the form where a, b R, over R with addition and scalar multiplication defined by 1 a + c 1 a 1 ka, k. b + k / V. We claim that V is indeed a vector space with the given operations. Note first that ( V) is( closed ) under the addtion and scalar multiplication operations: for, V and k R, we have and k 1 c 1 a + c V b + 1 ka V. k 1

2 VS 1: Observe that VS : Note that ( ) 1 a + c b + 1 c + a d + 1 c 1 a 1 e f 1 ( defn. of addition) ( comm. of addition in R) 1 a + c 1 e b + f 1 ( ) 1 (a + c) + e (b + d) + f 1 ( ) 1 a + (c + e) b + (d + f) 1 + e d + f 1 ( ( 1 e f VS 3: The matrix V and acts as the zero vector: a a a 1 a VS 4: Given any V, its additive inverse is as 1 a 1 a VS 5: Observe that 1. VS 6: Let k, l R, we have 1 a 1 (kl)a (kl) ( defn. of scalar mult) (kl) 1 k(la) ( mult. assoc. in R) k(lb) 1 1 la k l ( ) 1 a k l ( defn. of addition) ( addition again) ( add. assoc. in R) ))

3 VS 7: We have ( ) k VS 8: We have (k + l) 1 a + c k ( addition.) b + ( ) 1 k(a + c) k(b + d) 1 1 ka + kc ( dist. in R) kb + k 1 ka 1 kc k k 1 a 1 c k k 1 (k + l)a (k + l) 1 ka + la kb + l 1 ka 1 la k l ( ) ( 1 a k l ( dist. in R) ( addition.) )) ( 1 a (c) The set V of all positive real numbers over R with addition and scalar multiplication defined by x y xy, a x x a. We show that V is indeed a vector space with the given operations. Note first that if x, y V and a R, we have x y xy V, a x x a V so V is closed under addition and scalar multiplication. VS 1: We have VS : Note that x y xy yx ( mult. comm. in R) y x (x y) z (xy) z (xy)z x(yz) ( mult. assoc. in R) x (yz) x (y z) 3

4 VS 3: Observe that 1 V and 1 x 1x x. VS 4: For any x V, note that x 1 V so that x x 1 xx 1 1. VS 5: Note that 1 x x 1 x. VS 6: Let a, b R. We then have VS 7: Note that VS 8: We have (ab) x x ab (x b ) a ( exponents in R) a (x b ) a (b x) a (x y) a (xy) (xy) a x a y a ( exponents in R) (x a ) (y a ) (a x) (a y) (a + b) x x a+b x a x b ( exponents in R) (x a ) (x b ) (a x) (b x) (d) The set V of solutions of the differential equation f (t) 4f(t) t, t R over R with standard addition and scalar multiplication. Observe that V is not closed under addition: if f 1, f V, then for all t R so that f 1 (t) 4f 1 (t) t, f (t) 4f (t) t (f 1 + f ) (t) 4(f 1 + f )(t) (f 1 (t) 4f 1 (t)) + (f (t) 4f (t)) t t. We conclude that V is not a vector space with the given operations. 4

5 (e) The set V of invertible matrices with real entries over R with standard addition and scalar multiplication Observe that V is not closed under addition: we have, V but is not invertible. We conclude that V is not a vector space with the given operations.. By definition, every field F has a multiplicative identity, an element e such that e x x for every element x F. What is the multiplicative identity for R? Prove that the multiplicative identity is unique for any given field. The multiplicative identity for R is the number 1 as 1 x x for all x R. To show that the identity is unique, let e and e be two identities. Consider then the product e e. By thinking of e as an identity, we have e e e. Likewise, thinking of e as an identity leads to e e e so that e e. Thus, the identity is unique. Tuesday Aug Prove that the set of matrices with zero trace form a subspace of M n n (F ). Does the same hold for matrices with zero determinant? Let T be the set of matrices with zero trace. As M n n (F ) is a vector space over F and T is its subset, we merely need to check three properties: the matrix Z consisting only of zero entries evidently has zero trace so Z T. let A, B T ; it follows then that tr(a) tr(b) 0. Note then that tr(a + B) (A + B) ii i1 A ii + i1 let A T and k F ; we have tr(a) 0 so that B ii tr(a) + tr(b) 0. i1 tr(ka) (ka) ii k i1 A ii ktr(a) 0. i1 We conclude that T is a subspace of M n n (F ). The same cannot be said however about the set of matrices with zero determinant as it is not closed under addition. As an example, let A be the diagonal matrix with A 11 0 and A ii 1 for i,..., n and let B consist only of zeros except for B Then, det(a) det(b) 0 but det(a + B)

6 4. Let B(R) be the set of all bounded functions on R (A function f is bounded if there exists M such that f(x) M for all x. Thus sin(x) is bounded on R but e x is not). Prove that B(R) is a subspace of F(R, R), the set of all functions from R to R. As F(R, R) is a vector space and B(R) is its subset, we just need to check the following three properties: the function z 0 is clearly bounded (as z(x) 0 < 1 for all x) so z R. let f, g B(R). Then there exist M, N such that f(x) M and g(x) N for all x R. Note then that, by the triangle inequality (f + g)(x) f(x) + g(x) f(x) + g(x) M + N for all x R; thus, (f + g) is bounded and hence in B(R). let f B(R) and a R. Observe then that for all x R so af B(R). (af)(x) af(x) a f(x) a M We conclude that B(R) is a subspace of F(R, R). 5. Let W 1 and W be subspaces of a vector space V. Prove that W 1 + W W if and only if W 1 is a subspace of W. Suppose first that W 1 is a subspace of W. Let t W 1 + W ; there then exist w 1 W 1 and w W such that t w 1 + w. As W 1 is a subspace of W, it follows that w 1 W as well and hence t w 1 + w W so that W 1 + W W. As we also have W W 1 + W, we conclude that W 1 + W W. Conversely, suppose that W 1 + W W ; we want to show that W 1 is a subspace of W. Let w 1 W 1 and w W ; then, w 1 + w W 1 + W. As W 1 + W W, there exists some t W such that w 1 + w t w 1 t + ( w ) W. We conclude that W 1 W and, in particular, that W 1 is a subspace of W. 6. Let v 1 (0, 1) and v (1, 1) and define W 1 {tv 1 : t R} and W {tv : t R}. Also, let V R over R with standard operations. (a) Show that W 1 and W are subspaces of V. As W 1 and W are subsets of V which itself is a vector space, we just need to check the following three properties: (we treat both the spaces at the same time) 0 W i by setting t 0 in the definitions. let x, y W i. There then exist t, s R such that x tv i and y sv i so that x + y tv i + sv i (t + s)v i W i. 6

7 let x W i and a R. Note then that ax a(tv i ) (at)v i W i. We conclude that both W 1 and W are subspaces of V. (b) Show that V W 1 W. We need to show that (i) W 1 W {0} and (ii) W 1 + W V. For (i), note that if u W 1 W, then for some t, s R, we have u tv 1 and u sv so that tv 1 sv (0, t) (s, s). It follows that s 0 t 0 so u must be the zero vector. For (ii), let x (a, b) R. We want show that x w 1 + w for some w 1 W 1 and w W. Note that setting w 1 (0, b a) and w (a, a) accomplishes this as w i W i for i 1, and w 1 + w (0, b a) + (a, a) (a, b) v. As both (i) and (ii) hold, we conclude that V W 1 W. 7. Let E and O denote respectively the subsets consisting of all the even and odd functions in V : F(R, R). In the homework, you are supposed to show that both E and O are subspaces of V. Assuming that, prove that V E O. As in the previous problem, we just need to show that (i) E O {0} and (ii) E + O V. For (i), let f E O. Then, for any x R, we have f( x) f(x) and f( x) f(x) so that f(x) f(x) f(x) 0 f 0. For (ii), let f V. We need to show that f g + h where g E and h O. Define for all x R f(x) + f( x) f(x) f( x) g(x), h(x). Note then that g(x) + h(x) f(x) for all x. Furthermore, for any x R, we have g( x) f( x) + f(x) g(x) f( x) f(x) h( x) h(x). This shows that g E and h O and hence establishes (ii). V E O. We conclude that 7