Eigenvalue and Eigenvector Homework

Transcription

1 Eigenvalue and Eigenvector Homework Olena Bormashenko November 4, 2 For each of the matrices A below, do the following:. Find the characteristic polynomial of A, and use it to find all the eigenvalues of A. 2. State the algebraic multiplicity of each eigenvalue. 3. Find the fundamental eigenvectors of A, and state whether the matrix is diagonalizable. 4. If the matrix is indeed diagonalizable, check that P AP D holds. Here are the matrices in question: [ ] 2. A. Solution:. By definition, the characteristic polynomial of A is xi n A, and the notation for it is p A (x). Therefore, [ ] [ ] p A (x) xi n A 2 x x 2 x (x )(x ) (x ) 2 To find the eigenvalues, set p A (x) to. Thus, we get (x ) 2 x Therefore, the only eigenvalue is λ. 2. The algebraic multiplicity of the eigenvalue λ is the highest power of (x λ) that appears in the polynomial p A (x). Here, there s only λ, and the highest power of (x ) appearing in p A (x) is 2. Therefore, The algebraic multiplicity of λ is 2

2 3. Since there s only one eigenvalue, we only find the fundamental eigenvectors for it. We need to write E as all linear combinations of some set of vectors. We know that E λ is the set of solutions to (λi n A) x, so E is the set of solutions to (I 2 A) x. Now, [ ] [ ] [ ] 2 2 I 2 A The system (I 2 A) x corresponds to the following augmented matrix (which row reduces very simply): [ ] [ ] 2 R :/2 R This system clearly corresponds to Therefore, Thus, E { x A x x} {[ ] c c c c 2 } { [ ] c R c [ The only fundamental eigenvector is ] } c R (Of course, any scalar multiple of this is also correct.) Furthermore, for an n n matrix to be diagonalizable, it needs to have n different fundamental eigenvectors. Since here, we only have, and we would need 2, we conclude that the matrix is not diagonalizable. 4. Not applicable since the matrix is not diagonalizable A Solution:. Like above, p A (x) is defined to be xi n A. Thus, 3 3 p A (x) xi n A x x 3 3 x (x )((x + 5)(x 4) 3( 6)) 3 x 4 (x )(x 2 + x 2) (x ) 2 (x + 2) 2

3 To find the eigenvalues, set p A (x) to. Therefore, (x ) 2 (x + 2) x, 2 Therefore, the eigenvalues are λ and λ The algebraic multiplicity of the eigenvalue λ is the highest power of (x λ) that appears in the polynomial p A (x). Using the fact that p A (x) (x ) 2 (x + 2), we get that The algebraic multiplicity of λ is 2, and of λ 2 2 is 3. Since there are two eigenvalues, we need to figure out the eigenvectors for them separately. For each λ, we will be writing E λ as all linear combinations of some set of vectors. λ : From lecture, we have that E λ is the set of solutions to (λi n A) x. Therefore, E is the set of solutions to (I 3 A) x. Now, I 3 A Therefore, the system (I 3 A) x corresponds to the following augmented matrix: Row reducing, this becomes R2:R2 2R 3 3 R 3:R 3 R R :/3 R This clearly corresponds to the equation c 2 c 3 (and two copies of the equation ), and hence results in c c c 2 c 3 c 3 c 3 3

4 Therefore, E { x A x x} c c 3 c, c 3 c + c 3 c, Thus, the fundamental eigenvectors for λ are [,, ] T and [,, ] T. λ 2 2: Just like above, E λ is the set of solutions to (λi n A) x. This means that E 2 is the set of solutions to (( 2)I 3 A) x. Calculating, ( 2)I 3 A Therefore, the system (I 3 A) x corresponds to the following augmented matrix: Row reducing, this becomes R3:R3 R2 3 6 R :( /3) R R :/3 R 2 R :R +R 2 This corresponds to the equations: c c 3 c 2 2c

5 and therefore becomes c c 3 c 2 2c 3 c 3 c 3 Thus, we see that c 3 E 2 { x A x ( 2) x} 2c 3 c 3 c 3 2 This means that the only fundamental eigenvector for λ 2 2 is [, 2, ] T. We therefore conclude that The fundamental eigenvectors of A are, and 2 Since A is a 3 3 matrix, and we have 3 fundamental eigenvectors, we have a sufficient number. Therefore, the matrix is diagonalizable. 4. By definition, P is the matrix whose columns are the eigenvectors, and D is a diagonal matrix with the eigenvalues along the diagonal in the same order that we listed the eigenvectors. Therefore, here we have that P 2, D 2 Using the standard algorithm for the inverse, we calculate 2 R3:R3 R3 2 R :R +R 3 R 2:R 2+2R 3 R 3:( ) R

6 Therefore, P 2 Finally, checking that P AP D, 3 3 P AP D 2 as expected A 2. 3 Solution:. As usual, p A (x) xi n A. Thus, 3 p A (x) xi n A x 2 3 x 3 x 2 x 3 Expanding along the first row, x 3 x 2 (x 3) x 2 x 3 x 3 + x 3 (x 3)((x 2)(x 3) ) (x 3) (x 3)(x 2 5x + 5 ) (x 3)(x 2 5x + 4) (x 3)(x )(x 4) To find the eigenvalues, set p A (x) to. Therefore, (x 3)(x )(x 4) x, 3, 4 Hence, here the eigenvalues of A are λ, λ 2 3 and λ

7 2. The algebraic multiplicity of the eigenvalue λ is the highest power of (x λ) that appears in the polynomial p A (x). Since we know that p A (x) (x 3)(x )(x 4), The algebraic multiplicity of each of λ, λ 2 3 and λ 3 4 is 3. Let us figure out the eigenvectors for all the three eigenvalues separately. For each λ, we have to write E λ as all linear combinations of some set of vectors. λ : As we had above, E λ is the set of solutions to (λi n A) x. Therefore, E is the set of solutions to (I 3 A) x. Calculating, 3 2 I 3 A Hence, the system of equations given by (2I 3 A) x corresponds to the following augmented matrix: 2 2 Row reducing, 2 Swap R, R2 2 R 2:R 2+2R R 3:R 3+R 2 R 2:( ) R 2 R :R +R This clearly corresponds to the following equations: c c 3 c 2 2c 3 7

8 Solving, we get c c 3 c 2 2c 3 c 3 c 3 c 3 E { x A x x} 2c 3 c 3 c 3 2 Thus, the fundamental eigenvector for λ is [, 2, ] T. λ 2 3: Just like above, E λ is the set of solutions to (λi n A) x. This means that E 3 is the set of solutions to (3I 3 A) x. Calculating, 3 3 3I 3 A Therefore, the system (3I 3 A) x corresponds to the following augmented matrix: Row reducing, this becomes R3:R3 R Swap R, R 2 This corresponds to the system: R :R R 2 c + c 3 c 2 8

9 and therefore becomes c c 3 c 2 c 3 c 3 Thus, we get that c 3 E 3 { x A x 3 x} c 3 c 3 This means that the only fundamental eigenvector for λ 2 [,, ] T. 3 is λ 3 4: As before, E 4 is the set of solutions to (4I 3 A) x. Calculating, 4I 3 A Therefore, the system (4I 3 A) x corresponds to the following augmented matrix: 2 Row reducing to solve this, 2 R2:R2 R R 3:R 3 R 3 R :R R 2 This clearly corresponds to the following equations: c c 3 c 2 + c 3 9

10 Solving, we get Therefore, c c 3 c 2 c 3 c 3 c 3 E 4 { x A x 4 x} c 3 c 3 c 3 c 3 Thus, the fundamental eigenvector for λ 3 4 is [,, ] T. We therefore conclude that The fundamental eigenvectors of A are 2, and Since A is a 3 3 matrix, and we have 3 fundamental eigenvectors, this suffices. Therefore, A is diagonalizable. 4. By definition, P is the matrix whose columns are the eigenvectors, and D is a diagonal matrix with the eigenvalues along the diagonal in the same order that we listed the eigenvectors. Thus, from the derivations above, we get P 2, D 3 4 Finding the inverse, we see 2 R2:R2 2R R 3:R 3 R Swap R, R 2 R 2:(/2) R /2 /

11 R :R +R 2 R 3:R 3 2R 2 R 3:( /3) R 3 R :R R 3 /2 /2 /2 / /2 /2 /2 /2 3 /2 /2 /2 /2 /3 /3 /3 /6 /3 /6 /2 /2 /3 /3 /3 Therefore, /6 /3 /6 P /2 / /3 /3 / Finally, checking whether the equality P AP D holds, P AP D as expected A.. Calculating p A (x) xi n A, we get 2 3 p A (x) xi n A x x x + x + (x + ) 3

12 To find the eigenvalues, set p A (x) to. Thus, we get (x + ) 2 x Therefore, the only eigenvalue is λ. 2. The algebraic multiplicity of the eigenvalue λ is the highest power of (x λ) that appears in the polynomial p A (x). Since p A (x) (x + ) 3, and we only have the one eigenvalue λ, we conclude The algebraic multiplicity of λ is 3 3. Let us find the fundamental eigenvectors for our one eigenvalue λ. We need to write E as all linear combinations of some set of vectors. Since E λ is the set of solutions to (λi n A) x, E is the set of solutions to (( )I 3 A) x. Now, ( )I 3 A Writing this as an augmented matrix and row reducing, R:R+3R2 This system clearly corresponds to R :( /2) R Therefore, E { x A x x} c Thus, c c c 2 c 3 c R c The only fundamental eigenvector is c R Furthermore, for an n n matrix to be diagonalizable, it needs to have n different fundamental eigenvectors. Since here, we only have, and we would need 3, we conclude that the matrix is not diagonalizable. 4. Not applicable since the matrix is not diagonalizable. 2