MATH1120-2023-S2-workshop-wk12-sols

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MATH1120, S2 2023 Workshop Solutions Wk 12 Eigenvalues and Eigenvectors 1. Let A = 4 0 1 2 3 2 - 1 0 2 . Which of the following are eigenvectors for A , and what are their corresponding eigenvalues? (a) (0 , 1 , 0) T (b) (1 , 0 , 1) T (c) ( - 1 , 0 , 1) T (d) (0 , 2 , 0) T (e) (0 , 0 , 0) T (a) 4 0 1 2 3 2 - 1 0 2 0 1 0 = 0 3 0 = 3 0 1 0 Yes, this is an eigenvector with eigenvalue 3. (b) 4 0 1 2 3 2 - 1 0 2 1 0 1 = 5 4 - 1 No, not an eigenvector. (c) 4 0 1 2 3 2 - 1 0 2 - 1 0 1 = - 3 0 3 = 3 - 1 0 1 Yes, this is an eigenvector with eigenvalue 3. (d) 4 0 1 2 3 2 - 1 0 2 0 2 0 = 0 6 0 = 3 0 2 0 Yes, this is an eigenvector with eigenvalue 3. (e) 4 0 1 2 3 2 - 1 0 2 0 0 0 = 0 0 0 Even though the zero matrix as input gives the zero matrix as output, it is not allowed to count as an eigenvector. (Its eigenvalue would not be uniquely defined.) Wow, so far all the eigenvalues of this matrix are 3, and we have found only two linearly independent eigenvectors. Is this all? (Actually, yes, as you could test with the methods below.)
MATH1120, S2 2023 Workshop Solutions Wk 12 2. Find all eigenvalues and eigenvectors of the following matrices: (a) A = 1 1 - 2 4 (b) B = 1 - 1 1 1 2(a) First find eigenvalues: 0 = det( A - λI ) = det 1 - λ 1 - 2 4 - λ = (1 - λ )(4 - λ ) - ( - 2)(1) = 4 - 5 λ + λ 2 + 2 = λ 2 - 5 λ + 6 = ( λ - 2)( λ - 3) Therefore, eigenvalues are λ 1 = 2 , λ 2 = 3. Next, find eigenvectors for each eigenvalue. 1 - λ 1 1 | 0 - 2 4 - λ 1 | 0 = - 1 1 | 0 - 2 2 | 0 ' 1 - 1 | 0 0 0 | 0 Let x 2 = t . Then x 1 - t = 0 x 2 = t Therefore x 1 = t x 2 = t So that x 1 x 2 = t t So, eigenvectors with λ = 2 are of the form x = t 1 1 . Similarly, eigenvectors with λ = 3 are of the form x = s 1 2 .
MATH1120, S2 2023 Workshop Solutions Wk 12 2(b) First find eigenvalues: 0 = det( B - λI ) = det 1 - λ - 1 1 1 - λ = (1 - λ )(1 - λ ) - ( - 1)(1) = 1 - 2 λ + λ 2 + 1 = λ 2 - 2 λ + 2 Therefore, eigenvalues are λ = 2 ± 4 - 8 2 = λ = 2 ± 2 - 1 2 = 1 ± i . Write λ 1 = 1+ i and λ 2 = 1 - i . Next, find eigenvectors for each eigenvalue. 1 - λ 1 - 1 | 0 1 1 - λ 1 | 0 = 1 - (1 + i ) - 1 | 0 1 1 - (1 + i ) | 0 = - i - 1 | 0 1 - i | 0 1 - i | 0 0 0 | 0 Let x 2 = t . Then x 1 - it = 0 x 2 = t Therefore x 1 = it x 2 = t So that x 1 x 2 = it t So, eigenvectors with λ = 1 + i are of the form x = t i 1 . Similarly, eigenvectors with λ = 1 - i are of the form x = s i - 1 . Continued over page with equivalent right answer. 2(b) Continued. Note, the right answer may also be expressed in different ways including: Eigenvectors with λ = 1 + i are of the form x = t 1 - i . Eigenvectors with λ = 1 - i are of the form x = s 1 i . 3. Let B = 0 3 0 1 0 - 1 0 2 0 . Find an invertible 3 × 3 matrix P such that P - 1 BP = D is a diagonal matrix.
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