MATH 113 HOMEWORK 1: SOLUTIONS

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1 MATH 3 HOMEWORK : SOLUTIONS Problems From the Book: Axler Chapter, problems, 4, 8, 9, and 3. Solution to. Multiplying the numerator by its congruence directly gives a simplified expression of the ratio of two complex numbers: a + bi = a bi (a + bi)(a bi) = a bi a 2 + b 2. So just take c = a a 2 + b 2, d = b a 2 + b 2. Solution to 4. If a = 0 then we are done. If a 0, then multiplying both sides of the equation av = 0 by /a gives a (av) = a 0. The associativity of scalar product in vector space shows that the left hand side of the equation is precisely v = v. So we get v = 0, which ends the proof. Solution to 8. Suppose U, U 2,... is a collection of subspaces of V. We wish to show U U 2 is also a subspace of V. Let u, v be two arbitrary elements and a F. Since U i is a subspace of V for each i, we have 0 U i. So 0 Suppose u, v Then for each i, u, v U i. Since U i is a subspace of V, we have u + v U i. Thus u + v Suppose u i= U i and a F. Then for each i, u U i. Because U i is a subspace, we have au U i. Thus au Because i= U i is a nonempty subset of V which is closed under addition and scalar multiplication, it is a linear subspace of V. Solution to 9. Suppose U, W are subspaces of V such that U W is a subspace as well. Assume for the sake of contradiction that U W and W U. Then there exists u U, u / W and w / U, w W by definition. Then u, w U W so (by linearity in subspaces) u + w U W. Thus it is in either u + w U or u + w W. Without loss of generality we can suppose that u + w U (it s the same argument for W ). Then u U, u + w U so that w U, a contradiction. Hence our original assumption must have been false, so either U W or W U. Conversely, if U W then U W = W which is clearly a subspace, and if W U then U W = U which is also a subspace by assumption. Proof of 3. False. Counterexample: Let V be any nonzero vector space, let U = {0} and U 2 = V and W = V. Then U + W = V = U 2 + W, but U U 2. Of course there are many other examples.

2 2 MATH 3 HOMEWORK : SOLUTIONS Question. Let F be R or C and let X be any set. Let F X denote the set of all functions f : X F. Define addition and scalar multiplication on F X so that if f, g F X and a F then (f + g)(x) = f(x) + g(x), (af)(x) = a(f(x)). (a) Show that F X is a vector space under these operations. (b) Which of the following sets of functions from R to R are subspaces of R R? Justify your answers. {f : R R : f is continuous at all x R} {f : R R : f(x) 0 for all x R} {f : R R : f is differentiable at all x R} Proof. (a): We verify the axioms individually. First, note that F X is indeed closed under addition and scalar multiplication: addition of functions is obviously still a well-defined function, and same for scalar multiplication of functions. Commutativity: Let f, g F X, then for any x, (f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x); hence f + g = g + f for all x X, so they are identical functions in F X. Associativity: Let f, g, h F X. Similar to the above argument, ((f + g) + h)(x) = (f + g)(x) + h(x) = (f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x)) = (f + (g + h))(x). Additive identity: Consider the zero function z : X F defined asz(x) = 0 for all x X. Then for any f F X, Then (f + z)(x) = f(x) + z(x) = f(x) + 0 = f(x). Additive inverse: For any function f F X, consider the function n f F X defined by Multiplicative identity: Let f F X, then n f (x) := (f(x)). (f + n f )(x) = f(x) + ( f(x)) = 0 = z(x). (f)(x) = f(x) = f(x). Multiplicative associativity: Let f F X and a, b F, then (a(bf))(x) = a((bf)(x)) = a(bf(x)) = (ab)f(x) = ((ab)f)(x). Distributive property : Let a F, f, g F X, (a(f + g))(x) = a(f + g)(x) = a((f(x) + g(x)) = af(x) + ag(x) = (af)(x) + (ag)(x) = ((af) + (ag))(x). Distributive property 2: Let a, b F, f F X : ((a + b)f)(x) = (a + b)f(x) = af(x) + bf(x) = (af)(x) + (bf)(x) = ((af) + (bf))(x). This finishes all the axioms and shows that F X is a vector space. (b): The first and third are subspaces; the second is not. The second is not, since it is not closed under scalar multiplication: the constant function I : I(x) = for all x is nonnegative for all x but its negative n I (x) fails to be. To show that the first and the third are subspaces, we verify the three axioms. Additive identity: the zero function z : z(x) = 0 for all x is clearly continuous and differentiable. Closed under addition: If f, g R R are continuous (alt. differentiable), then f + g is as well (facts from high-school calculus). Closed under scalar multiplication: If f R R and a R then af is (continuous/differentiable) if f is (continuous/differentiable), again by high-school calculus.

3 MATH 3 HOMEWORK : SOLUTIONS 3 Question 2. Define E to be the set E = {[x] : x R, x > 0}. Define addition and scaler multiplication on E as follows: [x] + [y] = [x y] for [x], [y] E, and a[x] = [x a ] for [x] E, a R. Show that E with the above operations is a vector space over R. Proof. We verify each of the axioms as above. First, note that E is indeed closed under addition and scalar multiplication: if x, y > 0 then xy > 0 as well; and for any a R, x a > 0 is well-defined and positive. Commutativity: For any [x], [y] E, we have Associativity: For any [x], [y], [z] E we have [x] + [y] = [x y] = [y x] = [y] + [x]. ([x] + [y]) + [z] = [x y] + [z] = [(xy)z] = [x(yz)] = [x] + [yz] = [x] + ([y] + [z]). Additive identity: For the element [] E, we have [x] + [] = [x] = [x]. Additive inverse: For any [x] E, the element [/x] E satisfies Multiplicative identity: For any [x] E, we have [x] + [/x] = [x /x] = []. [x] = [x ] = [x]. Multiplicative associativity: For any [x] E, a, b R, we have Distributive property : Distributive property 2: a(b[x]) = a([x b ]) = [(x b ) a ] = [x ab ] = (ab)[x]. a([x] + [y]) = a[xy] = [(xy) a ] = [x a y a ] = [x a ] + [y a ] = a[x] + a[y]. (a + b)[x] = [x a+b ] = [x a x b ] = [x a ] + [x b ] = a[x] + b[x]. Question 3. Let U,..., U m be subspaces of a vector space V. Let W = U U m. Show that W is the smallest subspace of V that contains all of the U i. In particular, show that: (a) W U i for all i. (b) If W is a subspace of V so that W U i for all i, then W W. Proof. (a): By definition, W = {u + + u m u i U i for each i}. Fix any i, i m. and any u U i. Then since 0 U j for all j i, it follows that u = u W. This shows that u U i implies that u W, i.e. that U i W. Since this is true for any i we are done. (b): Suppose W is a subspace of V such that W U i for all i. Pick any vector w W ; we want to show that w W as well. By definition w can be written as w = u + + u m, u i U i. Note that u U W and u 2 U 2 W so by closure under addition, u + u 2 W. Similarly u 3 W, u 4 W, etc., so by adding these vectors one-by-one, we eventually end up with u + + u m W, i.e. w W. This shows that for any w W, w W, which implies by definition that W W as desired.

4 4 MATH 3 HOMEWORK : SOLUTIONS Question 4. Consider the following subspaces of R 3 : U = {(x, 0, 0) : x R}, U 2 = {(0, y, 0) : y R}, U 3 = {(0, 0, z) : z R}, U 4 = {(w, 0, w) : w R}. Describe the following subspaces: U 2 + U 4 U + U 3 U + U 3 + U 4 U + U 2 + U 3 + U 4 Which of the above are all of R 3? For which of the above are the sums also direct sums? Solution. (a): By definition, the sum of given subspaces are: U 2 + U 4 = {(w, y, w) : w, y R} U + U 3 = {(x, 0, z) : x, z R} U + U 3 + U 4 = {(x + w, 0, z + w) : x, z, w R} = {(x, 0, z) : x, z R} U + U 2 + U 3 + U 4 = {(x + w, y, z + w) : x, y, z, w R} = {(x, y, z) : x, y, z R} = R 3 The first and the second summation follows directly from definition of addition of sets. To see the third and fourth one, we need to be careful that the two sets {(x+w, 0, z+w) : x, z, w R} and {(x, 0, z) : x, z R} are actually equal. Indeed, {(x, 0, z) : x, z R} {(x + w, 0, z + w) : x, z, w R} is natural by setting w = 0. And for any x, z, w R, we define x = x + w, z = z + w, so that (x + w, 0, z + w) = (x, 0, z ) {(x, 0, z ) : x, z R} = {(x, 0, z) : x, z R}. This gives {(x + w, 0, z + w) : x, z, w R} = {(x, 0, z) : x, z R}. And the expression for U + U 2 + U 3 + U 4 is obtained similarly. (b): We need to tell whether the sums are direct sums. Let s first recall that a sum of linear subspaces V = U + U U k is a direct sum if each element v V can be written uniquely as a sum u + u u k, where u j U j. As a result, if the intersection of two subspaces U and U 2 of R 3 is {(0, 0, 0)}, then their sum U + U 2 is a direct sum. To see this, suppose v = u + u 2 = u + u 2 be two ways to write v V as a sum of elements in U and U 2. Then we have u u = u 2 u 2. Since u u belongs to U and u 2 u 2 belongs to U 2, these two equal elements must lie in U U 2 = {(0, 0, 0)}, or we have u = u and u 2 = u 2. This criterion immediately gives us two direct sums among all: because U U 3 = U 2 U 4 = {(0, 0, 0)}, the sum U 2 + U 4 and U + U 3 must be direct sum. And the rest two are not direct sum. For example, we have and (, 0, ) = (, 0, 0) + (0, 0, ) + (0, 0, 0) = (0, 0, 0) + (0, 0, 0) + (, 0, ) (, 0, ) = (, 0, 0) + (0, 0, 0) + (0, 0, ) + (0, 0, 0) = (0, 0, 0) + (0, 0, 0) + (0, 0, 0) + (, 0, ) in the third and fourth sum. There are, of course, infinitely many examples to choose. Question 5. Prove or provide a counter-example to the following statement: U, U 2, U 3 of a vector space V that U (U 2 + U 3 ) = (U U 2 ) + (U U 3 ). Solution. The statement if false generally. Here is a possible counter-example. Take V = R 3 and U = {(x, y, 0) : x, y R}, U 2 = {(x, y, y) : x, y R}, U 3 = {(0, 0, z) : z R}. For any three subspaces Clearly U, U 2, U 3 are subspaces of V. Now we have U 2 + U 3 = {(x, y, y + z) : x, y, z R} = {(x, y, z) : x, y, z R} = R 3, so the left hand side is U (U 2 + U 3 ) = U = {(x, y, 0) : x, y R}, a linear subspace of V. However, U U 2 = {(x, 0, 0) : x R},

5 MATH 3 HOMEWORK : SOLUTIONS 5 and U U 3 = {(0, 0, 0)}. So the right hand side is (U U 2 ) + (U U 3 ) = {(x, 0, 0) : x R} + {(0, 0, 0)} = {(x, 0, 0)}, an one-dimensional subspace of V. Now (0,, 0) is in the former subspace but not in the latter one. We have U (U 2 + U 3 ) (U U 2 ) + (U U 3 ). A little remark: in general we do have U (U 2 + U 3 ) (U U 2 ) + (U U 3 ), but the other direction of inclusion is not true. Question 6. Approximately how much time did you spend working on this homework?