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M346 First Midterm Exam Solutions, September 21, 2004 1. Let V be the subspace of R4 defined by the equation x1 + x2 + x3 + x4 = 0. a) Find the dimension of V . V is the null space of the rank-1 matrix ( 1 1 1 1 ), and so has dimension 4-1=3. b) Find a basis for V . [Any basis will do, but the simpler your answer, the easier part (c) will be. Be sure that each of your vectors really is in V , and that they are linearly independent] −1 1 , Treat x1 as the constrained variable and the others as free: 0 0 −1 −1 0 0 . , 1 0 1 0 x2 x 3 . Note that L takes V to V , and can be viewed as an c) Let L(x) = x4 x1 operator on V . Find the matrix [L]B , where B is the basis you found in part (b). = −b3 , L(b2 ) = b3 − b1 and L(b3 ) = −b1 , the matrix is Since L(b1 ) 0 1 0 0 1 0 . Of course, if you picked a different basis in (b) then you −1 −1 −1 would get a different matrix in (c). µ ¶ µ ¶ 5 3 2 2. In R , consider the basis b1 = , b2 = . 3 2 a) Find the change-of-basis matrices PEB and PBE , where E is the standard basis. µ ¶ µ ¶ 5 3 2 −3 −1 PEB = ([b1 ]E [b2 ]E ) = . PBE = PEB = . 3 2 −3 5 ¶ µ 13 , find [v]B . b) If v = −2 1 2 −3 13 32 [v]B = PBE [v]E = = . You can check that v is −3 5 −2 −49 indeed equal to 32b1 − 49b2 . µ ¶ µ ¶ x1 2x2 c) Let L = . Find [L]E and [L]B . By inspection, [L]E = x x + x 2 1 2 µ ¶ µ ¶ 0 2 −12 −7 . Then we compute [L]B = PBE [L]E PEB = . 1 1 22 13 3. Consider the coupled first-order differential equations µ ¶µ ¶ µ ¶ dx1 = x1 +2x2 dt dx2 = 2x1 + x2 dt Define the new variables y1 (t) = x1 (t) + x2 (t), y2 (t) = x1 (t) − x2 (t). a) Rewrite the system of equations completely in terms of y1 and y2 . (That is, express dy1 /dt and dy2 /dt as functions of y1 and y2 .) dy1 /dt = 3y1 , dy2 /dt = −y2 . This implies that y1 (t) = e3t y1 (0) and y2 (t) = e−t y2 (0). b) Given the initial conditions x1 (0) = 1, x2 (0) = 0, find x1 (t) and x2 (t). y1 (0) = 1 + 0 = 1 and y2 (0) = 1 − 0 = 1, so y1 (t) = e3t and y2 (t) = e−t , so x1 (t) = (e3t + e−t )/2 and x2 (t) = (e3t − e−t )/2. 4. Let V =R3 [t], and let L : V → V be defined by L(p)(t) = p0 (t) + 2p00 (t). a) Find [L]E , where E = {1, t, t2 , t3 } is the standard basis. By taking derivatives, we find that L(b1 ) = 0, L(b2 ) = 1 = b1 , L(b3 ) = 2t + 4 = 4b1 +2b2 , L(b4 ) = 3t2 + 12t = 12b2 + 3b3 , so the matrix of L is 0 1 4 0 0 0 2 12 . 0 0 0 3 0 0 0 0 b) What is the dimension of the kernel of L? What is the dimension of the range of L? Since the matrix has 3 pivots, L has rank 3, so the kernel has dimension 4 − 3 = 1 and the range has dimension 3. c) Find a basis for the kernel of L. {1} = {b1 }. d) Find a basis for the range of L. There are many correct answers. One is 2 {1, t, t2 }. Another, more directly from the columns of [L]E is {1, 4 + 2t, 12 + 3t2 }. Note that the basis vectors are elements of V (that is, functions), not columns of numbers. Their COORDINATES are columns of numbers, and form a basis for the column space of [L]E . 5. True of False? Each question is worth 4 points. You do NOT need to justify your answers, and partial credit will NOT be given. 1 0 2 0 0 1 1 0 For (a) and (b), suppose that a 4×4 matrix A row-reduces to . 0 0 0 1 0 0 0 0 a) The null space of A is the span of (−2, −1, 1, 0)T . TRUE. x3 is the only free variable, x1 = −2x3 , x2 = −x3 , and x4 = 0. 0 0 1 0 0 1 , , and . b) The column space of A is the span of 1 0 0 0 0 0 FALSE. The column space is the span of the first, 2nd and 4th columns of A, which could be almost anything. For (c) and (d), suppose that L :R2 [t] → M2,2 is a linear transformation, and that B = [L]EE is the matrix of L relative to the standard bases for R2 [t] and M2,2 . c) If B row-reduces to something with 3 pivots, then L is 1–1. TRUE. In that case L would have rank 3. Since R2 [t] is 3-dimensional, that would make the kernel 0-dimensional, so L is 1–1. 1 µ ¶ 3 1 3 d) If is in the range of L, then is in the column space of B. 4 4 7 7 TRUE. The coordinates of a vector in the range give a vector in the column space of the matrix. e) R3 is the internal direct sum of the x1 -x2 and x1 -x3 planes. FALSE. Those two subspaces overlap on the x1 axis. 3