HW11-solutions - calles(dac3476 HW11 gilbert(54425 This...

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calles (dac3476) – HW11 – gilbert – (54425) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points IF the matrix A = b 2 1 1 4 B is diagonalizable, i.e. , A = PDP - 1 with P invertible and D diagonal, which oF the Fol- lowing is a choice For D ? 1. A is not diagonalizable correct 2. D = b 3 0 0 3 B 3. D = b 3 0 0 3 B 4. D = b 3 0 0 2 B 5. D = b 3 0 0 2 B Explanation: Since det [ A λI ] = b 2 λ 1 1 4 λ B = 1 + (2 λ )(4 λ ) = 9 6 λ + λ 2 , the eigenvalues oF A are the solutions oF 9 6 λ + λ 2 = (3 λ ) 2 = 0 , i.e. , λ = 3 , 3. On the other hand, when λ = 3, rreF( A λI ) = rreF b 1 1 1 1 B = b 1 1 0 0 B , so x 2 is the only Free variable. Thus the eigenspace Nul( A 3 I ) has dimension 1. But then, when λ = 3, geo mult A ( λ ) < alg mult A ( λ ) . Consequently, A is not diagonalizable . 002 10.0 points An n × n matrix can be diagonalizable, but not invertible. True or ±alse? 1. ±ALSE 2. TRUE correct Explanation: Consider the 3 × 3 triangular matrix A = 5 8 1 0 0 7 0 0 2 Because A is triangular, its eigenvalues are the entries along the diagonal, i.e. , λ = 5 , 0 , 2. Since these are distinct, A is diagonalizable. On the other hand, one oF its eigenvalues is zero, so A is not invertible (or note that det[ A ] = 0 because det[ A ] = 5(0)( 2) = 0 is the product oF the diagonal values oF A , so A is not invertible). ThereFore, an n × n matrix A can be diagonalizable, but not invertible. Consequently, the statement is TRUE . 003 10.0 points IF the matrix A = b 1 4 1 2 B is diagonalizable, i.e. , A = PDP - 1 with P invertible and D diagonal, which oF the Fol- lowing is a choice For P ?
calles (dac3476) – HW11 – gilbert – (54425) 2 1. P = b 1 4 2 1 B 2. P = b 1 1 1 4 B 3. P = b 2 4 1 1 B 4. P = b 1 4 1 1 B correct 5. A is not diagonalizable Explanation: Since det [ A λI ] = b 1 λ 4 1 2 λ B = 4 (1 λ )(2 + λ ) = λ 2 + λ 6 , the eigenvalues of A are the solutions of λ 2 + λ 6 = ( λ + 3)( λ 2) = 0 , i.e. , λ = 3 , 2. Thus A is diagonalizable because the eigenvalues of A are distinct, and A = PDP - 1 with P = [ v 1 v 2 ] where v 1 and v 2 are eigenvectors correspond- ing to λ 1 and λ 2 respectively. To determine v 1 and v 2 we solve the equation A x = λ x . A x = λ x , λ = 3: b 1 4 1 2 Bb x 1 x 2 B = b x 1 + 4 x 2 x 1 2 x 2 B = 3 b x 1 x 2 B , which can be written as x 1 + 4 x 2 = 3 x 1 , x 1 2 x 2 = 3 x 2 , i.e. , x 1 = x 2 . So one choice of v 1 is v 1 = b 1 1 B . A x = λ x , λ = 2: b 1 4 1 2 Bb x 1 x 2 B = b x 1 + 4 x 2 x 1 2 x 2 B = 2 b x 1 x 2 B , which can be written as x 1 + 4 x 2 = 2 x 1 , x 1 2 x 2 = 2 x 2 , i.e. , x 1 = 4 x 2 . So one choice of v 2 is v 2 = b 4 1 B . Consequently, A = PDP - 1 with P = b 1 4 1 1 B . 004 10.0 points If an n × n matrix A is diagonalizable, then A has n distinct eigenvalues.

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