# HW 10 - HW 10 gilbert This print-out should have 22...

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HW 10 gilbert 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points If the matrix is diagonalizable, i.e., A = PDP1 with P invertible and D diagonal, which of the following is a choice for D? 1. 2. A is not diagonalizable correct 3.4.5.Explanation:Since ,the eigenvalues of A are the solutions of 9 6λ + λ2 = (3 λ)2 i.e., λ = 3, 1.FALSE 2.TRUE correctExplanation:Consider the 3 × 3 triangular matrix Because A is triangular, its eigenvalues are the entries along the diagonal, i.e., λ = 5, 2. Since these are distinct, 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 matrix A can be diagonalizable, but not invertible. Consequently, the statement is .TRUE
= 0,
3. On the other hand, when λ = 3, rref(AλI) = rref,so x2 is the only free variable. Thus the eigenspace Nul(A3I) has dimension 1. But then, when λ = 3, geo multA(λ) < alg multA(λ).Consequently, .002 10.0 points An n×n matrix can be diagonalizable, but not invertible. True or False?
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A is
× n
003 10.0 points If the matrix A is not diagonalizable
HW 10 gilbert is diagonalizable, i.e., A = PDP1 with invertible and D diagonal, which of the following is a choice for P1.correct2.3.4. A is not diagonalizable 5. Explanation:Since ,the eigenvalues of A are the solutions of λ2 + λ 6 = (λ + 3)(λ 2) = 0i.e., λ = 3, A are λ = 1, 2, P= 1 4
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2. Thus A is diagonalizable because the eigenvalues of A are distinct, and A has n distinct eigenvalues. True or False?
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x. A are λ = 1, 2, P= 1 4
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x2 ,
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= 0.Thus the eigenvalues of A are λ = 1, 2, P= 1 4
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HW 10 gilbert 3
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