True or False 1 TRUE 2 FALSE correct Explanation If an n n matrix has n

# True or false 1 true 2 false correct explanation if

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True or False? 1. TRUE 2. FALSE correct Explanation: If an n × n matrix has n distinct eigenvalues, it is diagonalizable, but the converse does not necessarily have to be true. For example, when A = 1 3 3 3 5 3 3 3 1 , then det[ A λI ] = ( λ 1)( λ + 2) 2 = 0 . Thus the eigenvalues of A are λ = 1 , 2 , 2. Now rref( A I ) = 1 0 1 0 1 1 0 0 0 , so Nul( A I ) = s 1 1 1 : s in R .
moreno (jcm5423) – HW11 – gilbert – (54425) 3 On the other hand, rref( A + 2 I ) = 1 1 1 0 0 0 0 0 0 , so Nul( A + 2 I ) = s 1 1 0 + t 1 0 1 : s, t in R . But then A is diagonalizable because it has 3 linearly independent eigenvectors 1 1 1 , 1 1 0 , 1 0 1 . Since the eigenvalue λ = 2 is repeated, how- ever, A does not have distinct eigenvalues. Consequently, the statement is FALSE . 005 10.0 points The eigenvalues of the matrix A = 2 0 2 1 3 2 0 0 3 are λ = 2 , 3 , 3. If A is diagonalizable, i.e. , A = PDP - 1 with P invertible and D diagonal, which of the following is a choice for P ? 1. P = 1 0 2 1 1 0 0 0 1 2. P = 1 0 2 1 1 0 0 0 1 3. P = 1 0 2 1 1 0 0 0 1 correct 4. A is not diagonalizable 5. P = 1 0 2 1 1 0 0 0 1 Explanation: We ±rst determine the eigenspaces corre- sponding to λ = 2 , 3: λ = 2 : since rref( A 2 I ) = rref 0 0 2 1 1 2 0 0 1 = 1 1 0 0 0 1 0 0 0 , there is only free variable and Nul( A 2 I ) = s 1 1 0 : s in R . Thus Nul( A 2 I ) has dimension 1, and v 1 = 1 1 0 is a basis for Nul( A 2 I ). λ = 3 : since rref( A 3 I ) = rref 1 0 2 1 0 2 0 0 0 = 1 0 2 0 0 0 0 0 0 , there are two free variables and Nul( A 3 I ) = s 0 1 0 + t 2 0 1 : s, t in R . Thus Nul( A 3 I ) has dimension 2, and v 2 = 0 1 0 , v 3 = 2 0 1 ,
moreno (jcm5423) – HW11 – gilbert – (54425) 4 form a basis for Nul( A 3 I ). Consequently, A is diagonalizable because v 1 , v 2 , v 3 are linearly independent, and A = PDP - 1 with P = [ v 1 v 2 v 2 ] = 1 0 2 1 1 0 0 0 1 . 006 10.0 points Every n × n matrix A having eigenvectors v 1 , v 2 , . . ., v n can be diagonalized. True or False? 1. TRUE 2. FALSE correct Explanation: An n × n matrix A can be diagonalized, i.e. written as A = PDP - 1 for some invert- ible matrix P and diagonal matrix D , if and only if A has linearly independent eigenvectors v 1 , v 2 , . . ., v n . Consequently, the statement is FALSE . 007 10.0 points If A is a diagonalizable n × n matrix, then A is invertible. True or False? 1. FALSE correct 2. TRUE 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.