# 1 false correct 2 true explanation a non zero vector

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1. FALSE correct 2. TRUE Explanation: A non-zero vector x is an eigenvector of A if and only if A x = λ x for some scalar λ . Now A x = 3 6 7 3 2 7 5 6 4 1 2 2 = 5 13 1 negationslash = λ 1 2 2 for any choice of scalar λ . Consequently, the statement is FALSE . 011 10.0points If the matrix A = bracketleftbigg 2 1 1 4 bracketrightbigg is diagonalizable, i.e. , A = PDP - 1 with P invertible and D diagonal, which of the fol- lowing is a choice for D ? 1. D = bracketleftbigg 3 0 0 3 bracketrightbigg 2. D = bracketleftbigg 3 0 0 2 bracketrightbigg 3. D = bracketleftbigg 3 0 0 2 bracketrightbigg 4. D = bracketleftbigg 3 0 0 3 bracketrightbigg 5. A is not diagonalizable correct Explanation: Since det [ A λI ] = bracketleftbigg 2 λ 1 1 4 λ bracketrightbigg = 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 bracketleftbigg 1 1 1 1 bracketrightbigg = bracketleftbigg 1 1 0 0 bracketrightbigg , 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 . 012 10.0points If the matrix A = bracketleftbigg 1 4 1 2 bracketrightbigg is diagonalizable, i.e. , A = PDP - 1 with P invertible and D diagonal, which of the fol- lowing is a choice for D ? 1. D = bracketleftbigg 1 0 0 2 bracketrightbigg 2. A is not diagonalizable 3. D = bracketleftbigg 1 0 0 2 bracketrightbigg 4. D = bracketleftbigg 3 0 0 2 bracketrightbigg 5. D = bracketleftbigg 3 0 0 2 bracketrightbigg correct Explanation:
huynh (lth436) – HW09 – gilbert – (57245) 6 Since det [ A λI ] = bracketleftbigg 1 λ 4 1 2 λ bracketrightbigg = 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 D = bracketleftbigg λ 1 0 0 λ 2 bracketrightbigg = bracketleftbigg 3 0 0 2 bracketrightbigg . 013 10.0points If the matrix A = bracketleftbigg 1 4 1 2 bracketrightbigg is diagonalizable, i.e. , A = PDP - 1 with P invertible and D diagonal, which of the fol- lowing is a choice for P ? 1. A is not diagonalizable 2. P = bracketleftbigg 1 1 1 4 bracketrightbigg 3. P = bracketleftbigg 1 4 1 1 bracketrightbigg correct 4. P = bracketleftbigg 1 4 2 1 bracketrightbigg 5. P = bracketleftbigg 2 4 1 1 bracketrightbigg Explanation: Since det [ A λI ] = bracketleftbigg 1 λ 4 1 2 λ bracketrightbigg = 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 .
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