HW2 - Due 01/30. Each answer must be mathematically justified. Don t forget your name.

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1 HW2 - Due 0/30 Each answer must be mathematically justified. Don t forget your name. Problem. Use the row reduction algorithm to find the inverse of the matrix 0 0, if it exists. Double check your answer. We use the row reduction algorithm. We have: Therefore the inverse is given by = To double check this answer we compute the matrix times its inverse: = As the right side is the identity matrix we are fine. Problem 2. Are the following matrices diagonalizable: 2 2 (a), (b), (c) ? (a) We compute the characteristic polynomial: λ 2 4 λ = ( λ)(4 λ) + 2 = 6 5λ + λ2. The polynomial 6 5λ + λ 2 has two distinct roots: 2 and 3. Therefore the matrix has two distinct eigenvalues. Since it is a 2 2 matrix, it is diagonalizable. (b) We compute the characteristic polynomial: 2 0 = ( λ)2. Therefore is an eigenvalue of algebraic multiplicity 2. However, Id 0 2 = =. 0 0

2 2 Since this is not the null matrix, is an eigenvalue of geometric multiplicity. Since the algebraic multiplicity is not equal to the geometric multiplicity, the matrix is not diagonalizable. (c) This is a symmetric matrix. Therefore it is diagonalizable by the spectral theorem. Problem 3. Diagonalize the following matrices (a) A = 7 24, (b) B = 2 7. Check your answers. (a) We first compute the characteristic polynomial: 7 λ λ = (7 λ)( 7 λ) + 48 = λ2. This has two roots: and. These are distinct thus the matrix is diagonalizable. As for the actual diagonalization, we find eigenvectors by looking at the nullspace of A Id 2 and A + Id 2, respectively: A Id 2 = is eigenvector with eigenvalue ; ] A + Id 2 = is eigenvector with eigenvalue. ] It follows that A = P DP where P is the matrix of the ordered eigenvectors and D is the matrix of the ordered eigenvalues: ( ) P =, D =, P 0 =. 4 To check this answer, we must perform a matrix multiplication to check that P DP = A. We have: P DP = = = So the answer is correct. (b) We observe that B has rank 2. Therefore, 0 is an eigenvalue of multiplicity 2. The sum of the eigenvalue of B is equal to the trace, hence the last eigenvalue must

3 be 3. We now find eigenvectors: B 0 Id 3 = 0 0 and 0 are eigenvectors with eigenvalue 0. 0 B 3Id 3 = is eigenvectors with eigenvalue 0. It follows that A = P DP where P is the matrix of the ordered eigenvectors and D is the matrix of the ordered eigenvalues: P = 0 0, D = To check the answer, it suffices to verify that P DP = B or equivalently that P D = BP (this trick allows to check the answer without computing P ). We have: P D = = 3 3, 3 3 BP = 0 0 = So the answer is correct. Problem 4. Compute A 208, where A is given by Problem 3(a): 7 24 A =. 2 7 Then find (x 208, y 208 ), where (x n, y n ) are given by the recursion formula: { { xn+ = 7x n 24y n x0 =, y n+ = 2x n 7y n y 0 =. We use the diagonalization formula: A = P DP, where P and D were computed previously. Then, A 208 = A A... A A = P DP P DP... P DP P DP. Since P P = Id 2, all the terms P P in the above formula cancel. Hence, we are left with: A 208 = A A... A A = P D D... D DP = P D 208 P. 3

4 4 Now we can conclude: D 208 = ( ) 208 = 0 = Id 0 2, A 208 = P P = Id 2. Regarding (x 208, y 208 ), we observe that if xn v n = then y v n+ = A v n. (0.) n Thus v 208 = A 208 v 0 = v 0 and (x 208, y 208 ) = (, ). Problem 5. Find all values of a such that the matrix 2 a A = is diagonalizable with real eigenvalues. a 4 We see a as a parameter. We first look for the characteristic equation: 2 λ a det = (2 λ)(4 λ) + a a 4 λ 2 = λ 2 6λ + (8 + a 2 ). The discriminant of this equation is = 36 4(8 + a 2 ) = 4 4a 2. For A to have real eigenvalues, we need to have 0, which forces a. In the case a <, > 0 and the characteristic equation has two distinct roots that correspond to two different eigenvalues. In this case the matrix A is diagonalizable. The remaining case is a = or equivalently a = ±. In this case = 0 and we have a repeated root λ = 3. In this case A is diagonalizable if we have a repeated eigenvector, hence an eigenspace of dimension 2. To find the eigenspace we look at the nullspace of A 3Id, 0] in the cases a =, a = : For a = :, only one eigenvector. For a = : ], only one eigenvector. Hence if a = ± A is not diagonalizable. Problem 6. Test whether these quadratic forms are positive definite or negative definite: (a) q (x, x 2 ) = 2x 2 + x 2 2 4x x 2, (b) q 2 (x, x 2 ) = 4x 2 + 4x 2 2 8x x 2, (c) (optional) q 3 (x, x 2, x 3 ) = x 2 + 2x 2 2 4x x x 2 2x 2 x 3 + 4x x 3. (a) q is represented by the symmetric matrix 2 2 Q =. 2 We use the minors test: 2 > 0 and 2 ( 2) 2 = 2 < 0. Thus q is indefinite. (b) q 2 is represented by the symmetric matrix 4 4 Q 2 =. 4 4

5 We use the minors test: 4 > 0 and 4 4 ( 4) 2 = 0. Thus q 2 is positive semidefinite. (c) q 3 is represented by the symmetric matrix Q 3 = We use the minors test: > 0, 2 2 = > 0 and = = = 7 Thus q 3 is indefinite. Problem 7. Let A be a n n symmetric matrix such that A 2 5A + 6Id n = 0. Show that A has only positive eigenvalues. Since A is symmetric, it is diagonalizable, and we must show that all eigenvalues of A are positive. Let λ be one of them. There exists a vector v 0 such that A v = λ v. Therefore, A 2 v = A A v = A λ v = λ A v = λ 2 v. In addition, A 2 5A + 6Id n = 0. It follows that (A 2 5A + 6Id n ) v = 0 A 2 v 5A v + 6 v = 0 (λ 2 5λ + 6) v = 0. Since v 0, λ must satisfies the equation λ 2 5λ + 6 = 0. This equation has solutions 2 and 3, thus λ = 2 or 3. In particular λ > 0. Since λ is any eigenvalue of A, A has only positive eigenvalues (and is definite positive). 5