In class midterm Exam - Answer key

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1 Fall 2013 In class midterm Exam - Answer key ARE211 Problem 1 (20 points). Metrics: Let B be the set of all sequences x = (x 1,x 2,...). Define d(x,y) = sup{ x i y i : i = 1,2,...}. a) Prove that d is a metric for B. Ans: (Symmetry) d(x,y) = sup{ x i y i : i = 1,2,...} = sup{ y i x i : i = 1,2,...} = d(y,x) (Nonnegativity) Every element of the set is nonnegative ( y i x i 0 for all i). The supremum of a non-empty set of non-negative numbers is non-negative (sup{ x i y i : i = 1,2,...} > 0). (Zero) d(x,x) = sup{ x i x i : i = 1,2,...} = sup{ 0 : i = 1,2,...} = 0 (Triangle Inequality) d(x,z)+d(z,y) = sup{ x i z i : i = 1,2,...} +sup{ z i y i : i = 1,2,...} sup{ x i z i + z i y i : i = 1,2,...} sup{ (x i z i )+(z i y i ) : i = 1,2,...} = sup{ x i y i : i = 1,2,...} = d(x,y) b) Now, let F be the set of all x B such that sup{ x i : i = 1,2,...} 1. c) Show that F is closed and bounded. (Note: a set of sequences F is bounded if there exists b R such that for all elements of the set, the distance between that element and the zero sequence is less than b.) Ans: (Bounded) Let 0 be a sequence of zeros. d(x,0) = sup{ x i 0 : i = 1,2,...} = sup{ x i : i = 1,2,...} 1 by construction. So the set is bounded by any number greater than 1. (Closed) Consider a convergent sequence x n is contained in the set F. Need to show the limit of this convergent sequence is in F (i.e., lim x n = x F ). For j = 1,2,...,limx n j = x j. We need to show that since each x n j [ 1,1], x j [ 1,1]. Assume, for purposes of contradiction, that limx n j = x j > 1. Let ε = x j 1 2. There exists an N N such that for all n > N, x n j x j < ε = x j 1 2. Which means for all n > N, x n j > 1, a contradiction. So x j 1 A similar proof with ε = 1 x j 2 gives you x j 1. Having shown x j [ 1,1], we know limx n = x F, so F is closed.

2 2 d) For n N, let F n = {x B s.t. x i 1, for 1 i n}. Provide a contra-positive proof that F n N Fn. Ans: For an arbitrary sequence x, assume x / n N F n. Then there exists some ˆn N where x / Fˆn. That means for some i < ˆn, x i > 1. Therefore x / F. e) Is F compact? Why or why not? Ans: F is not compact. Let B be the set of all bounded sequences y = (y 1,y 2,...). For each x F, let O(x) = {y B : d(x,y) < 1}. Note that O = xo(x) is an open cover of F. Now, define the sequences x (n) as follows, { x (n) x (n) = i = 1 if n = i = 1 if n i x (n) i For example, x (1) = (1, 1, 1,...) and x (2) = ( 1,1, 1, 1,...). Clearly, x (n) F for all n because d(x (n),0) = 1. Now show by contradiction that x (n),x (m) cannot belong to the same O(x) if n m. Assume x (n),x (m) O(x) and n m. d(x (n),x (m) ) d(x (n),x)+d(x,x (m) ) (by the triangle inquality) < 1+1 (because x (n),x (m) O(x)) = 2 A contradiction, because d(x (n),x (m) ) = 2 by construction. Therefore no distinct x (n),x (m) can belong to the same O(x). Since each O(x) can contain at most one sequence x (n), any finite collection of O(x) could not cover the set {x (n) : n = 1,2,...}. Since {x (n) : n = 1,2,...} is contained in F, no finite subset of O could cover F. Problem 2 (20 points). Sequences: Say whether each of the following statements is true or false. If the statement is true, provide a proof. If the statement is false, give a counter-example. Assume that X R is endowed with the Euclidean metric. a) Consider a sequence x n in X. i) If x n is convergent, then the sequence y n = x n+1 x n converges to 0. Ans: True. x n is a convergent sequence on a subset of R with the Euclidean metric, so x n is a Cauchy sequence. Therefore, for all ε > 0 there exists an N N such that for all n,m > N, d(x n,x m ) = x n x m < ε. Clearly, for all n > N, d(x n+1,x n ) = x n+1 x n < ε. So d(y n,0) = y n 0 = x n+1 x n < ε. Therefore, y n converges to zero. ii) If y n = x n+1 x n converges to 0, then x n is convergent. Ans: False. Let X = [0,1) (1,2] and x n = 1 1/n. Clearly X R. The sequence y n = x n+1 x n = 1 n(n+1) converges to zero, but x n does not converge in X.

3 b) Let x n and y n be sequences in X. Define the sequence z n in X to be the shuffled sequence {x 1,y 1,x 2,y 2,...,x n,y n,...}. Then z n is convergent if and only if x n and y n are both convergent with lim{x n } = lim{y n }. Ans: Assume z n converges some z. We have shown (PS 3, 1a) that if a sequence converges, then every subsequence of that sequence converges to the same limit. Both τ 1 (n) = 2n and τ 2 (n) = 2n + 1 are subsequences of z n. Since x n = τ 1 (n) and y n = τ 2 (n), both x n and y n converge to z. Assume x n and y n converge to z. Given ε > 0, there exists N 1,N 2 N such that for all n > N 1, d(x n,z) < ε and for all n > N 2, d(y n,z) < ε. Let N 3 = max{n 1,N 2 }. For all n > N 3, d(z n,z) < ε. c) EXTRA CREDIT: y n = x 1+x x n n. There exists a non-convergence sequence {x n } for which the sequence y n is convergent. Ans: These sequences are called Cesaro means. For a R, all sequences of the form x n = a( 1) n converge to zero. Also, all sequences of the form { 0 if n is even x n = a if n is odd converge to a 2. 3 Problem 3 (20 points). Hemi-continuity: a) Consider the correspondence ξ : R R defined by: { {0}, x 0 ξ = { x} [0, x], x > 0 i) Sketch a graph of the correspondence ξ. Ans: ii) Is ξ upper-hemi continuous, lower-hemi continuous, both, or neither? Why?

4 4 Ans: First show the correspondence is upper-hemi continuous and lower-hemi continuous on (,0]. For x (,0] and any open set U with U ξ( x), there exists some ε > 0, where B d (0,ε) U. Since f(0) is continuous, there exists a neighborhood V of x, such that ξ(x) B d (0,ε) U for all x V. Next, show the correspondence is upper-hemi continuous over (0, ). Let x (0, ) and U be an open set such that ξ( x) = { x} [0, x] U. Then x, x are interior points of U. Therefore, there exists an ε > 0, where B d ( x,ε) U and B d ( x,ε) U. Since [0, x] U, [0, x + ε] U. And since the square root function is continuous, there exists a neighborhood V of x such that for any x V we have x B d ( x,ε), so [0, x) [0, x+ε) U. Similarly, for any x V, we have x B d ( x,ε) U, so ξ(x) U. Finally, show the correspondence is lower-hemi continuous over (0, ). Given {s n } s, need to construct a sequence t n t for any t ξ( s). If t = x or t = x, then t n = s n or t n = s n, respectively. If t [0, x), then: { 0 if sn < t t n = t if s n t b) Next, consider the correspondence ξ : (0,3] R defined by: [x 2,x), 0 < x < 1 ψ = [0,1), x = 1 (1,x 2 ), 1 < x 3 i) Sketch a graph of the correspondence ξ. Ans: ii) Is ξ upper-hemi continuous, lower-hemi continuous, both, or neither? Why?

5 Ans: The correspondence is not upper-hemi continuous. The set O = (0,1) is an open set in the codomain. The upper inverse image of O, ψ 1 (O), is the set (0,1] in the domain, which is not open. The correspondence is not lower-hemi continuous. The set O = B d ( 1 2,ε) for some ε < 1 2 is an open set in the codomain. The lower inverse image of O, ψ 1 (O), is the set ( 1 2 +ε, 1 2 +ε) {1} in the domain, which is not open. 5 Problem 4 (20 points). Bases: In the following, you need to justify your answer for full credit. You can get a fair bit of credit for carefully drawn pictures, providing they are carefully enough drawn. But not-graphical formalism would add value. There s a bit of duplication here, as you will see: think of it as a little gift. a) Suppose that S = {v 1,v 2 } is a basis for R 2, and consider v 3 R 2. Let S i denote the set in which v i is replaced by v 3, e.g., S 1 = {v 2,v 3 }. Under what condition on v 3 will i) exactly one of the S i s be a basis for R 2. Ans: The condition is that for some α 0, and some i = 1,2, v 3 = αv i. In this case S i will be a basis for R 2. For j i, S j will not be a basis: it is a (non-minimal) spanning set for the 1-dimensional subspace of R 2 that s spanned by v i. ii) at least one of the S i s be a basis set for R 2. Ans: The condition is that v 3 0. In this case, either v 3 is collinear with one of the v i s in which case the previous answer applies. If it s collinear with neither then both S 1 and S 2 will be bases for R 2. b) Now suppose that S = {v 1,v 2,v 3 } is a basis for R 3, and consider v 4 R 3. Again, let S i denote the set in which v i is replaced by v 4, e.g., S 1 = {v 2,v 3,v 4 }. Under what condition on v 4 will i) exactly one of the S i s be basis sets for R 3. Ans: If for some α 0, and some i = 1,2,3, v 3 = αv i, then S i will be a basis for R 3. For any j i, S j is a non-minimal spanning set for a two-dimensional sub-space of R 3. ii) exactly two of the S i s be bases for R 3. Ans: If for i,j {1,2,3}, i j, v 4 a linear combination of v i and v j, but v 4 is collinear with neither of them, then iii) all three of the S i s be bases for R 3. Ans: There exists no pair i,j {1,2,3}, i j, such that v 4 belongs to the span of {v i,v j }. (Note that this condition is not satisfied for v 4 = 0.)

6 6 Problem 5 (20 points). Vector Spaces: For n {0} N, an n th degree polynomial on R is a function of the form f(x) = n k=0 a kx k, for k = 0,1,...n, with (a 0,a 1,...,a n ) R n+1 and a n 0. Let P(n) denote the set of such functions and let P denote the set of all polynomials on R, i.e., P = n P(n). a) Prove that P an infinite-dimensional vector space. (To establish the infinite-dimensionality of the space, you just have to show that there no finite set can be a basis.) Ans: Let f(x) = n k=0 a kx k, with a n 0. Let g(x) = m k=0 b kx k, with a m 0. Pick α,β R. Assume w.l.o.g. that n m. Then h(x) = αf(x)+βg(x) = m (αa k +βb k )x k +α k=0 n k=m+1 and hence belongs to P(n) for some n {0,...,n}, and hence to P. To prove that P is infinite dimensional, let S = {f 1,...f k } be any finite set of elements of P. That is f i P(n i ) for some i. Let n = max i=1,...k n i and note that no element of P(n+1) can be written as a linear combination of the elements of S. b) Is it true that for every n, P(n) a vector space? If so, provide a basis for P(n). If it is false for some subset of N s, explain why not for that subset. If it is true for some subset of N s, prove it for that subset. Ans: P(0) is a vector space with a singleton basis set {f} where f( ) = 1. For n > 0, P(n) is not a vector space. For example, consider n = 1 and f(x) = x. f P(1) but f +( f) is the zero function, which is not a member of P(n). c) Consider the set F = 2 0 {f P(n),, s.t. f(0) = f(1)}. i) Show that F is a vector space. a k x k Ans: Pick f,g F, α,β R, and consider h = αf+βg. Note first that h is a polynomial of degree 2. Second h(0) = αf(0)+βg(0) = αf(1)+βg(1) = h(1). Hence h F. ii) Prove that F has lower dimension than the space P = P(0) P(1) P(2). Ans: Let f(x) = x. Clearly f P(1) so that f P. But f(0) = 0 f(1) = 1. Hence f / F. Hence the dimension of F is less than that of P. iii) Find a basis set for F. (Hint: Suppose h(x) = 2 k=0 a kx k F. What does this imply about a 1 +a 2 +a 3?) Ans: A basis set is S = {f 0,f 2 }, where f 0 ( ) = 1. and f 2 (x) = x(x 1). Clearly f i P(i), for i = 1,2 and for each i, f i (0) = f i (1). Hence S F. Now consider h F, i.e., h = 2 k=0 a kx k, for some a 1 a 2,a 3. Since h(0) = h(1) it follows that a 0 +a 1 +a 2 = a 0, i.e., that a 1 = a 2. Hence we can write h(x) = a 0 f 0 +a 2 f 2. iv) What is the dimension of F? Ans: The dimension of F is equal to the number of elements in any one of its basis sets. Since we have proved that S is a basis for F, it follows that the dimension of F is 2.

7 7 Problem 6 (20 points). Rank (THIS QUESTION IS FOR EXTRA CREDIT): Consider a pair of 2 2 matrices such that AB is the zero matrix, but BA is not. a) Give an example of a pair A and B that satisfy the above condition. Hint: KISS (as in, keep it simply, stupid) Ans: Let A = 1 0 and B = b) Prove that both A and B have rank 1.. Note that A B = 1 0, while B A = 1 0. Ans: We will prove this by contra-positive. Assume that the conclusion does not hold, i.e., that that at least one of A or B has rank either zero or two. W.l.o.g., assume: i) A has rank 0. In this case, A =, so that BA = 0, so that A and B fail the second part of the assumed condition. ii) A has rank 2. In this[ case, ] A is invertible, so that for each b R 2, x = A 1 b. Hence AB = 0 implies B =. But in this case, BA = 0, so that, once again, A and B fail the second part of the assumed condition.