Midterm 1 Solutions Math 20250 Section 55 - Spring 2018 Instructor: Daren Cheng #1 Do the following problems using row reduction. (a) (6 pts) Let A = 2 1 2 6 1 3 8 17 3 5 4 5 Find bases for N A and R A, and determine the rank and nullity of A. Solution. The reduced row echelon form of A is A = 1 0 2 5 0 1 2 4. 0 0 0 0 Therefore the general solution to the system A x = 0 is. [x 1 x 2 x 3 x 4 ] T = [2x 3 5x 4 2x 3 4x 4 x 3 x 4 ] T From here we know that a basis for N A is = x 3 [2 2 1 0] T + x 4 [ 5 4 0 1] T. [2 2 1 0] T, [ 5 4 0 1] T. Going back to the reduced row echelon form of A, we see that only the first two columns have pivots. Thus the first two columns of A, i.e. [ 2 1 3] T and [1 3 5] T, form a basis for R A. We also find that rank(a) = 2 = nullity(a). (b) (5 pts) Determine whether the polynomials f 1 (t) = 1 t + t 2, f 2 (t) = t t 2 and f 3 (t) = 2 2t are linearly independent in P 2. Solution. Use the basis A = (1, t, t 2 ) on P 2. Then [f 1 ] A = [1 1 1] T, [f 2 ] A = [0 1 1] T, [f 3 ] A = [2 2 0] T. Thus, to check whether the given polynomials are linearly independent, we row reduce the following matrix: A = 1 0 2 1 1 2, 1 1 0 and find that its reduced row echelon form is the identity I 3. Hence the given polynomials are linearly independent.
#2 (6 pts) Let V = P 3 and let W = R 2. Define T : V W by letting T (f) = [f (1) f(1)] T. Find a basis for N T and determine the nullity and rank of T. Solution. Consider the basis A = (1, t, t 2, t 3 ) for V, and the standard basis, which we denote by B, for W. Then since T (t k ) = [k 1] T, for k = 0, 1, 2, 3, we know that [ ] 0 1 2 3 [T ] BA =. 1 1 1 1 After row reduction, we find its reduced row echelon form to be [ ] 1 0 1 2. 0 1 2 3 Thus, the null space of the matrix representation [T ] BA has a basis given by [1 2 1 0] T and [2 3 0 1] T, which correspond to f 1 (t) = 1 2t + t 2 and f 2 (t) = 2 3t + t 3 under our choice of basis for V. Thus f 1 and f 2 form a basis for N T. The nullity of T is then 2. Since dim V = 4, by the rank-nullity theorem, the rank of T is also 2.
#3 Let V be a finite-dimensional vector space over F. (a) (6 pts) Suppose v 1,, v r V are linearly independent in V. Prove that if span{ v 1,, v r } V, then there exists v r+1 V such that v 1,, v r+1 are linearly independent in V. Solution. Please look at the solutions to homework 1 for this problem. (b) (6 pts) Let W be a subspace of V. Prove that dim W dim V, and that if dim W = dim V then necessarily W = V. Solution. For convenience we let m = dim W and n = dim V. Suppose w 1,, w m is a basis for W, then in particular they are linearly independent in V. We learned through solving linear equations that in a finite-dimensional vector space, the size of a linear independent set of vectors cannot exceed the dimension of the space. In the present case, this implies that m n, and we are done with the first part. The second part concerns the case where m = n. Again let w 1,, w n be a basis for W. We claim that they span V. To see this, suppose by contradiction that they don t, then by part (a) there exists w n+1 V such that w 1,, w n+1 are independent in V. However, this is impossible since n + 1 > n. Therefore we must have V = span{ w 1,, w n } = W (the second equality follows from the fact that the w i s are a basis for W to begin with), and thus V = W as desired.
#4 Let V be a vector space over F and suppose T L(V, V ). (a) (4 pts) Suppose V is finite-dimensional. Prove that N T = { 0 V } if and only if R T = V. Solution. If N T = { 0 V }, then dim N T = 0, and thus by the rank-nullity theorem we have dim R T = dim V 0 = dim V. By part (b) of the previous problem, this implies that R T = V (recall that R T is a subspace of V ). Conversely, if R T = V, then obviously dim R T = dim V, so by the rank-nullity theorem we have dim N T = dim V dim V = 0, and thus N T = {0}. (b) (6 pts) Let V = C([0, 1]; R) (continuous real-valued functions on [0, 1]). Find a linear transformation T L(V, V ) such that N T = {0} (here 0 means the zero function), but that R T V. Proof. Such examples abound, but I ll only give one. Let T (f)(t) = tf(t) and note that if tf(t) = 0 for all t [0, 1], then f(t) = 0 for all t (0, 1]. By continuity, this gives f(t) = 0 for all t [0, 1], i.e. f = the zero function on [0, 1]. Therefore N T = {0}. On the other hand, by definition, T (f)(0) = 0f(0) = 0 no matter what f is. In other words, any function in V that does not vanish at the origin cannot be in R T. In particular, 1 / R T. We conclude that R T V.
#5 Let V = C 1 ([0, 1]; R) (continuously differentiable real-valued functions on [0, 1]) and let W = C([0, 1]; R). Define T : V W by letting T (f)(t) = f (t) f(t). (a) (6 pts) Prove that if f N T, then there exists a constant c R such that e t f(t) = c for all t [0, 1]. Solution. Note that ( e t f(t) ) = e t ( f (t) f(t) ) = e t T (f)(t). Thus if f N T, then by the above calculation and the fundamental theorem of calculus, we have e t f(t) = c for some constant c R. Consequently f(t) = ce t. (b) (5 pts) Use part (a) to show that dim(n T ) = 1. Proof. What we proved in part (a) amounts to the fact that N T span{e t }. On the other hand, it s easy to see that for any c R, the function ce t lies in N T, and so we also have the reverse inclusion span{e t } N T. In conclusion, N T = span{e t } and thus dim N T = 1.