The desired basis is 12 14 7 8 57 if we can find a

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Elementary Linear Algebra
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Chapter 6 / Exercise 36
Elementary Linear Algebra
Larson
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The desired basis is 12 14 , - 7 - 8 . 57. If we can find a basis B = ( v 1 , v 2 , v 3 ) such that the B -matrix of A is B = 1 0 0 0 1 0 0 0 - 1 , then A must be similar to 1 0 0 0 1 0 0 0 - 1 . Because of the entries in the matrix B , it is required that Av 1 = v 1 , Av 2 = v 2 and Av 3 = - v 3 . So, all we need for our basis is to pick independent v 1 , v 2 in the plane, and v 3 perpendicular to the plane. 58. a. Consider a linear relation c 1 A 2 v + c 2 Av + c 3 v = 0. Multiplying A 2 with the vectors on both sides and using that A 3 v = 0 and A 4 v = 0, we find that c 3 A 2 v = 0 and therefore c 3 = 0, since A 2 v = 0. 170
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Elementary Linear Algebra
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Chapter 6 / Exercise 36
Elementary Linear Algebra
Larson
Expert Verified
ISM: Linear Algebra Section 3.4 Therefore, our relation simplifies to c 1 A 2 v + c 2 Av = 0. Multiplying A with the vectors on both sides we find that c 2 A 2 v = 0 and therefore c 2 = 0. Then c 1 = 0 as well. We have shown that there is only the trivial relation among the vectors A 2 v, Av , and v , so that these three vectors from a basis of R 3 , as claimed. b. T ( A 2 v ) = A 3 v = 0 so [ T ( A 2 v )] B = 0 0 0 . T ( Av ) = A 2 v so [ T ( Av )] B = 1 0 0 . T ( v ) = Av so [ T ( v )] B = 0 1 0 . Hence, by Fact 3.4.3, the desired matrix is 0 1 0 0 0 1 0 0 0 . 59. First we find the matrices S = x y z t such that 2 0 0 3 x y z t = x y z t 2 1 0 3 , or 2 x 2 y 3 z 3 t = 2 x x + 3 y 2 z z + 3 t . The solutions are of the form S = - y y 0 t , where y and t are arbitrary constants. Since there are invertible solutions S (for example, let y = t = 1), the matrices 2 0 0 3 and 2 1 0 3 are indeed similar. 60. First we find the matrices S = x y z t such that 1 0 0 - 1 x y z t = x y z t 0 1 1 0 , or, x y - z - t = y x t z . The solutions are of the form S = y y - t t , where y and t are arbitrary constants. Since there are invertible solutions S (for example, let y = t = 1), the matrices 1 0 0 - 1 and 0 1 1 0 are indeed similar. 61. We seek a basis v 1 = x z , v 2 = y t such that the matrix S = [ v 1 v 2 ] = x y z t satisfies the equation - 5 - 9 4 7 x y z t = x y z t 1 1 0 1 . Solving the ensuing linear system 171
Chapter 3 ISM: Linear Algebra gives S = - 3 z 2 z 4 - 3 t 2 z t . We need to choose z and t so that S will be invertible. For example, if we let z = 6 and t = 1, then S = - 9 0 6 1 , so that v 1 = - 9 6 , v 2 = 0 1 . 62. We seek a basis v 1 = x z , v 2 = y t such that the matrix S = [ v 1 v 2 ] = x y z t satisfies the equation 1 2 4 3 x y z t = x y z t 5 0 0 - 1 . Solving the ensuing linear system gives S = z 2 - t z t . We need to choose both z and t nonzero to make S invertible. For example, if we let z = 2 and t = 1, then S = 1 - 1 2 1 , so that v 1 = 1 2 , v 2 = - 1 1 . 63. First we find the matrices S = x y z t such that p - q q p x y z t = x y z t p q - q p , or, px - qz py - qt qx + pz qy + pt = px - qy qx + py pz - qt qz + pt . If q = 0, then the solutions are of the form S = - t z z t , where z and t are arbitrary constants. Since there are invertible solutions S (for example, let z = t = 1), the matrices p - q q p and p q - q p are indeed similar. (If q = 0, then the two matrices are equal.) 64. If b and c are both zero, then the given matrices are equal, so that they are similar, by Fact 3.4.6.a. Let’s now assume that at least one of the scalars b and c is nonzero; reversing the roles of b and c if necessary, we can assume that c = 0.

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