MATH PRACTICE EXAM 1 SOLUTIONS

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1 MATH 2359 PRACTICE EXAM SOLUTIONS SPRING 205 Throughout this exam, V and W will denote vector spaces over R Part I: True/False () For each of the following statements, determine whether the statement is true or false, and provide justification (if it is true you should show that it is true in general, and if it is false it suffices to provide a single counterexample) The vector spaces need not be finite dimensional! (i) For any linear map T : V W, the image of the map, T (V ) = {w W w = T v for some v V } V is a vector subspace of the vector space W The statement is true Let T : V W be any linear map It suffices to show that for any w, y T (V ) and s R, the linear combination of the form w + sy is also in the image Since w, y T (V ), there exists u, v V such that w = T u and y = T v Thus, by linearity of T : V W w + sy = T u + st v = T (u + sv) = w + sy T (V ), so T (V ) W is a subspace, as claimed (ii) A set of vectors {u, v, w} V is a linearly dependent set if and only if one vector is a sum of the other two The statement is false A simple counter-example is easily furnished in R 2 : ( ) ( ) ( ) 0 2 u =, v =, w = 0 2 This is clearly linearly dependent since 2u + 2v w = 0 However, no pair of the vectors sums to the remaining vector (iii) Two distinct planes in R 3 either intersect in a line, or have no common points The statement is true Recall two plane equations are equivalent if and only if one is a multiple of the other Thus, if we assume distinct planes, we have two possibilities: (a) The planes arise as different level surfaces of the same linear function ax + by + cz, or (b) The planes arise from different linear functions, which are not proportional

2 2 SPRING 205 In case (a), we have a system of the form a b c d a b c d a b c d 2 d 2 d This system is inconsistent, so there can be no solution Thus there can be no common points in this case In case (b) we will have two pivots since the rows are independent, hence there will be one free variable, and thus the solution space is a line (iv) A linear map T : V W is surjective if and only if the kernel is nontrivial, ie ker T contains a nonzero v The statement is absolutely false The map T : V W sending every vector v V to the zero vector 0 W possess a nontrivial kernel for any vector space V of dimension, but clearly is not surjective Similarly, there exist bijective linear maps, called isomorphisms, for which the kernel is trivial and the map is surjective Bonus : If a map T : V V is injective, then it is necessarily surjective This statement is false, though any linear examples are infinite-dimensional (as we just learned about the rank-nullity theorem, a good exercise is to see why it implies for finite dimensional V, the statement holds for linear T ) Note that the question didn t say that T needs to be linear, so a simple example is the map f : R R given by f(x) = e x A simple linear example is the unique linear map of polynomials determined by replacing x by x 2 (ie ev x 2P(R) P(R)) It maps the infinite dimensional basis {, x, x 2, } to the basis of the subspace of even polynomials, given by even-degree monomials {, x 2, x 4, }, and hence cannot be surjective, since no odd polynomials are in the image The kernel is the zero polynomial, since any nonzero polynomial is either constant and unaffected or is sent to a higher degree polynomial 2 Part II In this part, please attempt 4 of the remaining 6 questions Clearly indicate which problems you wish to have graded (2) Let A, B Mat n n (R) be invertible n n matrices Show that (AB) = B A Recall that the inverse matrix of A has the property A A = I = AA Thus, we examine the matrix (AB) with the property that (AB) (AB) = I = (AB)(AB) Taking either one of these, we can show the result For example, if we take the first equation, we can multiply both sides by B on the right, to obtain (AB) (AB)B = B = (AB) A(BB ) = (AB) AI = (AB) A = B

3 MATH 2359 PRACTICE EXAM SOLUTIONS 3 Then multiplying both sides by A on the right, one obtains as was to be shown (AB) AA = B A = (AB) = B A, (3) Solve the following system of equations and express the solution(s) as a either a vector or linear combination of vectors: x 3y + 5z w = x + 3y + z 2w = 2x + 6z 3w = 0 6y + 4z + w = 2 The row operations for Gauss-Jordan are: R 2 R R 2, R 3 2R R 3, R 3 R 2 R 4, R 4 + R 2 R 3, 6 R 2 R 2, R + 3R 2 R This yields the reductions: / /3 /6 / Thus the system has solution x 0 y z = /3 0 + s w 0 3 2/3 0 + t 3/2 /6 0

4 4 SPRING 205 (4) Find a 2 2 matrix A Mat 2 2 (R) satisfying 2 2 A 0 2 = A Write A = Then = 2 0 x y 2x z w 2z 2 2 = = 2 2 We thus obtain equations y = 2, w =, 2x y = 2x + 2 = = x = 3/2, 2z w = 2z + = 2 = z = /2 Thus the matrix A is A = 3/2 2 /2 (5) Show that the following vectors are linearly independent: 0 0 3, 0 0, 3 5 We check that there are 3 pivots in the matrix whose columns are these vectors The row operations to do this are R 2 3R R 2, R 2 R 3, / 3R 2 R 2, R 2 + 3/ 5R 3 R 2

5 MATH 2359 PRACTICE EXAM SOLUTIONS 5 This gives the reduction / Thus, since we can put the matrix into RREF form with 3 pivots, the three vectors are linearly independent (6) The set of upper triangular 3 3 matrices UT 3 3 (R) := a a 2 a 3 0 a 4 a 5 a i R 0 0 a 6 is a vector space Calculate the dimension dim R UT 3 3 (R) by producing a basis of the space One basis, which is an analogue of the standard basis of R n, is given by the matrices having a one in precisely one of the entries on or above the diagonal, and zeroes elsewhere: B := 0 0, 0 0 Thus dim R UT 3 3 (R) = 6, (7) Show that the map of polynomials given by, 0 0 I : P(R) P(R), p(x) Ip(x) :=, p(t) dt, is a linear map, and by calculating ker I, show that it is not injective Is the map surjective? That this defines a linear map is clear: let p(x), q(x) P(R) be any two polynomials, and s R any real number Then it suffices to show that Ip + sq(x) = Ip(x) + siq(x)

6 6 SPRING 205 This follows immediately from properties of integrals: Ip + sq(x) = p(t) + sq(t) dt = p(t) dt + s = Ip(x) + siq(x) q(t) dt To figure out the kernel, consider the integral of a basis monomial: { x Ix n = t n dt = 2 ( )n+ x n+ = n+ xn+ if n even, n + 0 if n odd By linearity, we can determine the map entirely by its effect on the basis, hence we conclude that the kernel consists of all polynomials whose only powers appearing are odd Since the kernel is nontrivial, the map is not injective The map also fails to be surjective, since the only nonzero polynomials in the image can be expressed as integrals of even polynomials, and thus contain only odd powers by the above computation A remark: this is an interesting example: we ve essentially constructed an operator which projects one subspace (the subspace of even polynomials) onto a complementary subspace (the subspace of odd polynomials), while killing the subspace onto which we are projecting Applying the operator twice thus gives us the zero map; this is an example of a nilpotent order 2 linear map