Mathematics 6 Solutions for HWK Section Problem 9. Given T(x, y, z) = (x 9y + z,6x + 5y z) and v = (,,), use the standard matrix for the linear transformation T to find the image of the vector v. Note that the domain for T is R and the codomain is R. So we re expecting a matrix. Since T(,, ) = (, 6), T(,,) = ( 9,5), and T(,, ) = (, ), the standard matrix for T is ] 9 A = 6 5 and the standard coordinate matrix for T(v) is A In other words, T(v) = (5, 7). = ] 9 6 5 = ] 5. 7 Problem. Given that T : R R is the reflection through the origin, T(x, y) = ( x, y), and given v = (, ), (a) find the standard matrix A for the linear transformation T, (b) use A to find the image of the vector v, and (c) sketch the graph of v and its image. (a) Since T(, ) = (, ) and T(, ) = (, ), the standard matrix is A = ]. (b) T(v)] = Av] = ] ] = ] (c) See the sketch in the text. Page of 8 A. Sontag May,
Math 6 HWK Solns contd Problem 5. Given that T : R R is the counterclockwise rotation of 5 in R, and given v = (, ), (a) find the standard matrix A for the linear transformation T, (b) use A to find the image of the vector v, and (c) sketch the graph of v and its image. (a) T(,) = (cos 5,sin 5 ) = (, ) T(,) = (cos 5,sin 5 ) = (, ) A = ] (b) T(v)] = A v] = ] ] = ] T(v) = (, ) (c) See the text for a sketch. Problem 7. Given that T : R R is the reflection through the xy-coordinate plane in R, T(x, y, z) = (x, y, z), and given that v = (,, ), (a) find the standard matrix A for the linear transformation T, (b) use A to find the image of the vector v, and (c) sketch the graph of v and its image. (a) T(,,) = (,,), T(,,) = (,, ), and T(,,) = (,, ) so A = (b) T(v)] = Av = = so T(v) = (,, ).. (c) See the text for a sketch. Page of 8 A. Sontag May,
Math 6 HWK Solns contd Problem. Given that T : R R is the projection onto the vector w = (,) in R, T(v) = projwv, and given that v = (, ), (a) find the standard matrix A for the linear transformation T, (b) use A to find the image of the vector v, and (c) sketch the graph of v and its image. (a) (b) T(v)] = Av = T(, ) = proj (,) (,) = (,) T(, ) = proj (,) (,) = (,) A = 9 ] 9 ] ] = T(v) = 7 (,) ] = 7 7 ] (c) See the sketch in the text. Problem 5. Find the standard matrices for T = T T and T = T T, given T : R R, T (x, y) = (x y, x + y) T : R R, T (x, y) = (x, x y) ] The standard matrix for T is A =. ] The standard matrix for T is A =. In this situation matrix multiplication corresponds ] to composition of functions, so the standard matrix for T is A A = 5 ] and the standard matrix for T is A A =. 7 Page of 8 A. Sontag May,
Math 6 HWK Solns contd Problem 7. Given T : R R, T(x, y) = (x + y, x, y), v = (5, ) B = {(, ),(, )}, B = {(,,),(,,), (,, )} find T(v) by using (a) the standard matrix and (b) the matrix relative to B and B. (a) The standard matrix for T is A = so and T(v) = (9, 5,). ] 9 T(v)] = Av] = 5 = 5 (b) Denote the vectors in B, in the order given, as v, v. Similarly, let the vectors in B be called w, w, w. Then T(v ) = (,, ) = w w, T(v ) = (,, ) = w. Therefore the matrix for T, relative to B and B is A = Moreover, v = 5v + 9v. This gives ] T(v )] B T(v )] B =. T(v)] B = A v] B = ] 5 9 = 5 and consequently T(v) = 5w + w = (9,5, ), which agrees with the result found in (a). Page of 8 A. Sontag May,
Math 6 HWK Solns contd Problem. Given T : R R, T(x, y, z) = (x, x + y, y + z, x + z), v = (, 5,) B = {(,, ),(,, ),(,,)}, B = {(,,,),(,,, ),(,,,), (,,,)} find T(v) by using (a) the standard matrix and (b) the matrix relative to B and B. (a) The standard matrix for T is A = so so T(v) = (,,,). T(v)] = Av] = 5 = (b) Denote the vectors in B by v, v, v, and those in B by w, w, w, w. Then T(v ) = (,,, ) = w + w + w + w T(v ) = (,,, ) = w + w + w w T(v ) = (,,, ) = w + w + w + w The matrix for T relative to B and B is therefore A = Moreover, v = 9 v + v 8v. Therefore ] T(v )] B T(v )] B T(v )] B = T(v)] B = A v] B = 9 8 6 = which gives T(v) = 6w w w w = (,,,), which agrees with the result from (a). Page 5 of 8 A. Sontag May,
Math 6 HWK Solns contd Problem 5. Let T : P P be given by T(p) = xp. (In other words, if p(x) = c + c x + c x, then T(p) is the polynomial defined by (T(p))(x) = x(p(x)) = c x + c x + c x. Find the matrix of T relative to the bases B = {, x, x } and B = {, x, x, x }. Therefore the required matrix is T() = x, so T()] B =. T(x) = x, so T(x)] B =. T(x ) = x, so T(x )] B =. ] T()] B T(x)] B T(x )] B =. Problem 7. Let B = {, x, e x, xe x } be a basis of a subspace W of the space of continuous functions, and let D x be the differential operator on W. (In other words D x : W W is the linear transformation defined by D x (f) = f = the derivative function for the function f.) Find the matrix for D x relative to the basis B. D x () =, D x (x) =, D x (e x ) = e x, and D x (xe x ) = e x + xe x, so the required matrix is ] A = D x ()] B D x (x)] B D x (e x )] B D x (xe x )] B = Page 6 of 8 A. Sontag May,
Math 6 HWK Solns contd Problem 9. Use the matrix from Exercise 7 to evaluate D x (x xe x ). Let f be the function we wish to differentiate using Exercise 7. Then D x (x xe x )] B = D x (f)] B = Af] B = = This tells us that D x (f)(x) = e x xe x, exactly as we would expect from calculus. Problem 5. Let B = {, x, x, x } be a basis for P, and let T : P P be the linear transformation given by T(x k ) = t k dt. (a) Find the matrix A for T with respect to B and the standard basis for P. (b) Use A to integrate p(x) = 6 x + x. (a) T() = T(x ) = T(x ) = dt = x, T(x) = t dt = x, T(x ) = t dt = x t dt = x Give the name B to the standard basis for P : B = {, x, x, x, x }. Then the required matrix for T is ] A = T()] B T(x)] B T(x )] B T(x )] B = (b) The instructions are a little vague. Let s assume that what s wanted is to find T(6 x+x ). Then we have T(6 x + x )] B = A 6 6 x + x ] B = Thus T(6 x + x ) = 6x x + x. = 6. Page 7 of 8 A. Sontag May,
Math 6 HWK Solns contd Problem 55. Let T : M, M, be given by T(A) = A T. Find the matrix for T relative to the standard bases for M, and M,. Having consulted the text on p, and following the order suggested by Example 5, I ll take the standard basis for M, to be {M, M, M, M, M 5, M 6 }, where M, M, and M have zeros in all positions of the bottom row and all except one position of the top row and they have in the first, second, third positions, respectively of the first row. Then M, M 5, M 6 have zeros in all positions of the top row and two positions of the bottom row, with in the first, second, and third positions, respectively of the second row. (Yes, I m trying to avoid having to type in all those matrices.) I ll take the standard basis for M,, which I ll write as B = {N, N, N, N, N 5, N 6 }, to be arranged in like fashion: first let the s move across the first row, then across the second row, and finally across the third row, always from left to right. If you ordered these two bases differently, your representing matrix will come out different from mine, but your results should be consistent with mine once you take that difference into account. So here goes, finally. T(M ) = (M ) T = N, T(M ) = (M ) T = N T(M ) = (M ) T = N 5, T(M ) = (M ) T = N T(M 5 ) = (M 5 ) T = N, T(M 6 ) = (M 6 ) T = N 6 Therefore the matrix we want is. Page 8 of 8 A. Sontag May,