PROBLEM SET. Problems on Eigenvalues and Diagonalization. Math 3351, Fall Oct. 20, 2010 ANSWERS

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1 PROBLEM SET Problems on Eigenvalues and Diagonalization Math 335, Fall 2 Oct. 2, 2 ANSWERS

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3 Problem. In each part, find the characteristic polynomial of the matrix and the eigenvalues of the matrix by hand computation. A. A = [ ] Answer: We first compute A λi. [ 5 6 A λi = 2 3 ] [ λ ] [ 5 λ 6 = 2 3 λ ] The characteristic polynomial is the determinant of this matrix, so p(λ) = det(a λi) = (5 λ)( 3 λ) + (2)(6) = (5)(3) 5λ + 3λ + λ 2 + (2)(6) = 95 2λ + λ = 3 2λ + λ 2 Thus, p(λ) = λ 2 2λ 3 = (λ 3)(λ + ). The eigenvalues of A are the roots of the characteristic polynomial, so the eigenvalues are and 3. B. A = Answer: We have A λi = 2 λ 82 λ λ

4 Expanding along the first row (to take advantage of the ), we have λ 25 p(λ) = (2 λ) 4 6 λ + 82 λ 24 4 = (2 λ)[( λ)(6 λ) ( 25)4] + [ (82)4 24( λ)] = (2 λ)[ 66 + λ 6λ + λ 2 + ] + [ λ] = (2 λ)[λ 2 + 5λ + 34] + 24λ 64 = 2λ 2 + λ + 68 λ 3 5λ 2 34λ + 24λ 64 = 4 3λ 2 λ 3 Thus, p(λ) = 4 3λ 2 λ 3. We want to find the roots of this polynomial. The possible rational roots are the factors of 4, so the possibilities are ±, ±2 and ±4. It s easy to check that is a root. Long division then gives p(λ) = (λ )(λ 2 + 4λ + 4) = (λ )(λ + 2) 2, so the eigenvalues are and 2 ( 2 has multiplicity 2 are root of p(λ).) Problem 2. In each part, you are given a matrix A and the eigenvalues of A. Find a basis for each of the eigenspaces. Determine if A is diagonalizable and, if so, find an invertible matrix P and a diagonal matrix D so that P AP = D. A. The eigenvalues are and 2 and A = Answer: Use a calculator on this or you ll go insane. Let s start with the eigenvalue λ =. We want to find a basis of the eigenspace E( ), which is the same thing as the nullspace of the matrix A ( )I = A + I. We have A + I = The Reduced Row Echelon Form of A + I is 6 R = 2.. 2

5 We find the nullspace of this is the usual way. Let the variables be x, x 2, x 3. Then x and x 2 are leading variables, and x 3 is a free variable, say x 3 = α. The first row of R tells us that x + 6x 3 = = x = 6x 3 = x = 6α and the second row tells us that x 2 + 2x 3 = = x 2 = 2x 3 = 2α. Thus, the nullspace of R is parametrized by x 6 α x 2 = 2 α = α α x Thus, the eigenspace E( ) is one dimensional with basis vector 6 2. Next, consider the eigenvalue λ = 2. We compute that A 2I = The RREF of this is R = 2. We proceed as before. In this case, x is a leading variable and x 2 and x 3 are free variables. Say x 2 = α and x 3 = β. The top row of R tells us x 2x 2 + x 3 = = x = 2x 2 x 3 = 2α β. Thus, the nullspace of R is parametrized by x 2 α β 2 x 2 = α = α + β β x 3. 3

6 Thus, the eigenspace E(2) is two dimensional with basis 2,. We can put the bases of E( ) and E(2) together to get a basis of three dimensional space, so A is diagonalizable. To diagonalize it, we put the basis we ve found into a matrix P and make D the corresponding diagonal matrix with the eigenvalues on the diagonal. So, we can take 6 2 P = 2 D = 2. 2 Note that the diagonal entry in each column of D is the eigenvalue that goes with the corresponding column of P. (We could permute the columns of P, and arrange the eigenvalues in D to match, to get another solution of the problem.) Check with your calculator that P AP = D, or equivalently that A = P DP. B. The eigenvalues are and 2 and the matrix is A = Answer: First consider the eigenvalue λ =. We calculate that A + I = The RREF of this is R = 2. 4

7 The method for finding the nullspace of R is exactly as above. The result is that E( ) is one dimensional with basis 2. For eigenvalue λ = 2, we compute that 3 29 A 2I = The RREF of this is R = Let s work through the nullspace computation. There are leading entries in column 2 and column 3, so x 2 and x 3 are leading variables, and x is a free variable, say x = α. The second row of R tells us The first row tells us x + x 2 + x 3 = = x 3 =. x + x 2 + x 3 = = x 2 =. Thus, the nullspace of R is parametrized by x α x 2 = = α x 3. Thus, E(2) is one dimensional with basis vector. We only have two linearly independent eigenvectors, so we can not find a basis consisting of eigenvectors. Thus, we conclude that A is not diagonalizable. 5

8 C. The eigenvectors are and ± i, and the matrix is A = Answer: If you don t believe that some of the eigenvalues are complex, work out the characteristic polynomial on your calculator and use the function csolve or czeros to find the roots. For λ =, we can calculate that A I = and that the RREF of A I is 3/5 R = 4/5 In this case, x and x 2 are leading variables, and x 3 is a free variable, say x 3 = α. Reading up from the bottom of R we have Thus, we have x x 3 = = x 2 = 4 5 x 3 = 4 5 α x 3 5 x 3 = = x 2 = 3 5 α. x x 2 x 3 = 3 5 α 4 5 α α = α We conclude that E() is one dimensional with basis vector

9 We can avoid the mental agony of fractions if we observe the following fact: multiplying each vector in a basis by a nonzero constant yields another basis. In this case, we can multiply our basis vector by 5 to get another basis vector for E(). For eigenvalue λ = + i, we calculate that 3 i 4 5 A ( + i)i = 3 4 i i Fortunately, our calculator will handle complex numbers (use the symbol i right above the catalog key for i = ). Thus, we can enter the matrix above in the calculator and find that the RREF is R = 7 + i 7 + i It s easy to find the nullspace of this by our usual method. We see that x and x 2 are leading variables and x 3 is free, say x 3 = α. The first row of R says x + ( 7/ + i/)x 3 = = x = (7/ i/)α Similarly, the second row says x 2 = (7/ i/)α.. Thus, the nullspace of R is parametrized by x (7/ i/)α x 2 = (7/ i/)α = α α x 3 7/ i/ 7/ i/. We conclude that E( + i) is one dimensional with basis vector 7/ i/ 7/ i/. 7

10 We can get rid of the fractions by multiplying this by, to get another basis vector 7 i 7 i. for E( + i). We re left with the eigenvalue λ = i, which is the complex conjugate of + i. We don t need to do any calculation. Since A is real, conjugation gives a one-to-one correspondence between E( + i) and E( i) that sends bases to bases. Thus, to get a basis of E( i), we just take the conjugate of our basis vector for E( + i). Thus, E( i) is one dimensional with basis vector 7 + i 7 + i Since we get a basis of eigenvectors, we conclude that A is diagonalizable and we can take 3 7 i 7 + i P = 4 7 i 7 + i 5 D =. + i i. I invite you to calculate P DP to see that it really works. 8