Lecture 13: Complex Eigenvalues & Factorization

Lecture 13: Complex Eigenvalues & Factorization

Lecture 13: Complex Eigenvalues & Factorization Consider the rotation matrix 0 1 A = . (1) 1 ¡0 · ¸ To …nd eigenvalues, we write ¸ 1 A ¸I = ¡ ¡ , ¡ 1 ¸ · ¡ ¸ and calculate its determinant det (A ¸I) = ¸2 + 1 = 0. ¡ We see that A has only complex eigenvalues ¸= p 1 = i. § ¡ § Therefore, it is impossible to diagonalize the rotation matrix. In general, if a matrix has complex eigenvalues, it is not diagonalizable. In this lecture, we shall study matrices with complex eigenvalues. Since eigenvalues are roots of characteristic polynomials with real coe¢cients, complex eigenvalues always appear in pairs: If ¸0 = a + bi is a complex eigenvalue, so is its conjugate ¸¹0 = a bi. ¡ For any complex eigenvalue, we can proceed to …nd its (complex) eigenvectors in the same way as we did for real eigenvalues. Example 13.1. For the matrix A in (1) above, …nd eigenvectors. ¹ Solution. We already know its eigenvalues: ¸0 = i, ¸0 = i. For ¸0 = i, we solve (A iI) ~x = ~0 as follows: performing (complex) row operations to¡get (noting that i2 = 1) ¡ ¡ i 1 iR1+R2 R2 i 1 A iI = ¡ ¡ ¡ ! ¡ ¡ ¡ 1 i ! 0 0 · ¡ ¸ · ¸ = ix1 x2 = 0 = x2 = ix1 ) ¡ ¡ ) ¡ x1 1 1 0 take x1 = 1, ~u = = = i x2 i 0 ¡ 1 · ¸ ·¡ ¸ · ¸ · ¸ is an eigenvalue for ¸= i. For ¸= i, we proceed similarly as follows: ¡ i 1 iR1+R2 R2 i 1 A + iI = ¡ ! ¡ = ix1 x2 = 0 1 i ! 0 0 ) ¡ · ¸ · ¸ 1 1 1 0 take x = 1, ~v = = + i 1 i 0 1 · ¸ · ¸ · ¸ is an eigenvalue for ¸= i. We note that ¡ ¸¹0 = i is the conjugate of ¸0 = i ¡ ~v = ~u (the conjugate vector of eigenvector ~u) is an eigenvector for ¸¹0. This observation can be justi…ed as a general rule as follows. by taking conjugate on equation (A + ¸ I) ~u = 0 = (A + ¸ I) ~u = 0 0 ) 0 = (A + ¸ I) ~u = 0 = A¹ (¸ I) ~u = 0 = A ¸¹ I ~u = 0. ) 0 ) ¡ 0 ) ¡ 0 ¡ ¢ ³ ´ ¡ ¢ ¡ ¢ Remark. In general, if ~u is an eigenvector associated with ¸, then ~u, the conjugate of ~u, is an eigenvector associated with ¸¹, the conjugate of ¸. Therefore, for each pair, ¸, ¸¹ , of eigenvalue, we only need to compute eigenvectors for ¸. The eigenvectors for ¸¹ can be obtained easily by taking conjugates. ¡ ¢ Though A is not diagonalizable in the classic sense, we can still simplify it by introducing a term called "block-diagonal" matrix. Example 13.2. For the matrix A in (1) above that has complex eigenvalues, we proceed to choose P and D as follows: pick one complex eigenvalue and its eigenvector 1 1 0 ¸0 = i, ~u = = i i 0 ¡ 1 ·¡ ¸ · ¸ · ¸ separate their real part and imaginary part (Re (a + bi) = a, Im (a + bi) = b) : 1 0 Re (¸ ) = 0, Im (¸ ) = 1, Re (~u) = , Im (~u) = . 0 0 0 1 · ¸ ·¡ ¸ We then assemble D and P as Re (¸) Im (¸) 0 1 D = = , Im (¸) Re (¸) 1 0 ·¡ ¸ ·¡ ¸ 1 0 P = [Re (~u) , Im (~u)] = . 0 1 · ¡ ¸ Note that D is not really a diagonal matrix. We can verify AP = P D : 0 1 1 0 0 1 AP = = 1 ¡0 0 1 1 0 · ¸ · ¡ ¸ · ¸ 1 0 0 1 0 1 P D = = . 0 1 1 0 1 0 · ¡ ¸ ·¡ ¸ · ¸ 2 In general, if a matrix A has complex eigenvalues, it may be similar to a block-diagonal matrix B, i.e., there exists an invertible matrix P such that AP = P B, where B has the form C1 0 ... 0 0 C ... 0 B = 2 , P = [P , P , ..., P ] (2) 2 ... ... ... ... 3 1 2 r 6 0 0 ... Cr7 6 7 C is either a real eigenvalue4¸ ( 1 1 matr5ix), or C is a 2 2 matrix (called a block), k k £ k £ corresponding to a complex eigenvalue ¸k = Re (¸k) + i Im (¸k) , in the form Re (¸ ) Im (¸ ) C = k k , (3) k Im (¸ ) Re (¸ ) ·¡ k k ¸ and P is assembled accordingly: if Ck is an real eigenvalue, the corresponding column Pk in P is an eigenvector (real); if Ck is a 2 2 block as in (3) associated with a complex eigenvalue ¸k, £ then Pk consists of two columns: Pk = [Re (~uk) , Im (~uk)] where ~uk is a complex eigenvector for ¸k (note that both Re (~uk) and Im (~uk) are real vectors). Example 13.3. Some examples of block diagonal matrices (2): block matrix from v.s. normal matrix form. Example 1 1. 3 3 matrix with one diagonal entry and one 2 2 block: £ £ 2 0 2 0 0 2 3 = 0 2 3 , (¸ = 2, ¸ = 2 + 3i, ¸ = 2 3i) 0 1 2 3 2 3 2 3 20 3 23 ¡ ·¡ ¸ ¡ 4 5 4 5 2. 4 4 matrix with two diagonal entries and one 2 2 block: £ £ 1 0 0 1 0 0 0 ¡0 2 0 ¡0 2 0 0 2 3 = 2 3 , (¸1 = 1, ¸2 = 2, ¸3 = 4 + 2i, ¸4 = 4 2 0 0 4 2 ¡ ¡ 0 0 ¡ ¡ 6 2 4 7 6 0 0 2 47 6 ·¡ ¡ ¸7 6 ¡ ¡ 7 44 2i) 5 4 5 ¡ ¡ 3. 4 4 matrix with two 2 2 blocks: £4 2 £ 4 2 0 0 0 ¡2 4 ¡2 4 0 0 = , (¸ = 4+2i, ¸ = 4 2i, ¸ = 2+3i, 2·¡ ¡ ¸ 2 3 3 2¡0 ¡0 2 33 1 2 3 0 ¡ ¡ ¡ 6 3 2 7 6 0 0 3 27 6 ·¡ ¸7 6 ¡ 7 ¸4 =4 2 3i). 5 4 5 ¡ 1 The process of …nding a block diagonal matrix B in the form (2) and P such as P ¡ AP = B is called factorization. Note that if all eigenvalues are real, factorization = diagonalization. Factorization process: Step #1. Find all eigenvalues. 3 Step #2. For real eigenvalues, …nd a basis in each eigenspace Null (A ¸). For each pair of complex eigenvalues ¸ and ¸¹, …nd a (complex) basis in Null (A ¸¡) . Then, write complex eigenvectors in the basis in the form ~u = Re (~u) + i Im (~u) . ¡ Note that, the total number of such vectors must be equal to the dimension. Otherwise, it is not factorizable. Step #3. Assemble P and B as in (2): if C1 is a real eigenvalue, the corresponding column P1 is one of its eigenvectors in the basis; if C1 is a 2 2 block matrix as in (3), the £ corresponding P1 consists of two columns (Re (~u) , i Im (~u)) . Remark. Suppose that A is a upper (or lower) block-triangle matrix in the form B1 ... 0 B¤ ... ¤ A = 2 , 2 ... ... ... .¤.. 3 6 0 0 ... Br7 6 7 4 5 where each Bi is a square matrix (with same or di¤erent sizes). Then det (A) = det (B ) det (B ) det (B ) . 1 2 ¢ ¢ ¢ r det (A ¸) = det (B1 ¸) det (B2 ¸) det (Br ¸) . ¡ ¡ ¡ ¢ ¢ ¢ ¡ Therefore, eigenvalues of A = collection of all eigenvalues of B1, ..., Br. Example 13.4. Factorize 2 9 0 2 1 2 1 0 A = . 2¡0 0 3 0 3 6 0 0 1 17 6 ¡ 7 4 5 Solution. Step 1. Find eigenvalues. We notice that 2 ¸ 9 0 2 2 ¸ 9 ¡ ¡ 1 2 ¸ 1 0 1 2 ¸ ¤ A ¸I = 2 ¡ ¡ 3 = 2· ¡ ¡ ¸ 3 ¡ 0 0 3 ¸ 0 3 ¸ 0 ¡ 0 ¡ 6 0 0 1 1 ¸7 6 1 1 ¸ 7 6 ¡ ¡ 7 6 · ¡ ¡ ¸7 4 5 4 5 is an upper block-triangle. Thus, 2 ¸ 9 3 ¸ 0 det (A) = det ¡ det ¡ 1 2 ¸ ¢ 1 1 ¸ µ· ¡ ¡ ¸¶ · ¡ ¡ ¸ = [(2 ¸) (2 ¸) + 9] [(3 ¸) ( 1 ¸)] = 0. ¡ ¡ ¡ ¡ ¡ Hence [(2 ¸) (2 ¸) + 9] = 0 = ¸2 4¸+ 13 = 0 = ¸= 2 3i ¡ ¡ ) ¡ ) § or (3 ¸) ( 1 ¸) = 0 = ¸= 3, ¸= 1. ¡ ¡ ¡ ) ¡ 4 There are two real eigenvalues ¸1 = 3, ¸2 = 1, and one pair of complex eigenvalue ¹= 2+3i, and 2 3i, with real part and imaginary pa¡rt ¡ Re (¹) = 2, Im (¹) = 3. Step 2. We next …nd eigenvectors. For ¸1 = 3, we carry out row operation: 2 3 9 0 2 ¡1 2 3 1 0 A 3I = 2 ¡ ¡ 3 ¡ 0 0 3 3 0 ¡ 6 0 0 1 1 37 6 ¡ ¡ 7 4 5 19 1 9 0 2 R2 R1 R2 1 9 0 2 1 0 0 ¡ ¡ ! ¡ ¡ 5 1 1 1 0 R3 R4 0 10 1 2 2 1 3 = 2¡ ¡ 3 ! 2 ¡ ¡ 3 0 1 0 . 0 0 0 0 ! 0 0 1 4 ! 6 ¡5 7 ¡ 60 0 1 4 7 6 0 0 1 47 6 0 0 0 0 7 6 ¡ 7 6 ¡ 7 6 7 60 0 0 0 7 4 5 4 5 6 7 4 5 System became 19 x x = 0 1 ¡ 5 4 1 x2 x4 = 0 ¡ 5 x3 4x4 = 0. ¡ The eigenvalue is, by taking x4 = 5, 19 1 ~v = , ¸ = 3. 1 2203 1 6 5 7 6 7 4 5 For ¸2 = 1, we proceed as follows: ¡ 3 9 0 2 3 9 0 2 0 18 0 2 1 3 1 0 1 3 0 0 3R2+R1 R1 1 3 0 0 A + I = 2¡ 3 2¡ 3 ! 2¡ 3 0 0 4 0 ! 0 0 1 0 ! 0 0 1 0 6 0 0 1 07 6 0 0 0 07 6 0 0 0 07 6 7 6 7 6 7 4 5 4 5 4 5 R1/18 R1 0 1 0 1/9 1 0 0 1/3 ! ( 3) R1 + R2 R2 1 0 0 1/3 0 1 0 1/9 ¡ ! 2¡ ¡ 3 2 3 . ! 0 0 1 0 ! 0 0 1 0 6 0 0 0 0 7 60 0 0 0 7 6 7 6 7 4 5 4 5 5 We obtain that 1 x = x 1 ¡3 4 1 x2 = x4 ¡9 x3 = 0. By taking x4 = 9, ¡ 3 1 ~v2 = 2 3 , ¸2 = 1. 0 ¡ 6 97 6¡ 7 For ¹ = 2 + 3i, we proceed analogously:4 5 2 (2 + 3i) 9 0 2 ¡ 1 2 (2 + 3i) 1 0 A ¹I = 2 ¡ ¡ 3 ¡ 0 0 3 (2 + 3i) 0 ¡ 6 0 0 1 1 (2 + 3i)7 6 ¡ ¡ 7 4 5 R1 3iR2 R1 ¡ ! 3i 9 0 2 R4 R3/ (1 3i) R4 0 0 0 2 ¡ ¡ ¡ ! 1 3i 1 0 R2 R3/ (1 3i) R2 1 3i 0 0 = 2 ¡ ¡ 3 ¡ ¡ ! 2¡ ¡ 3 0 0 1 3i 0 ! 0 0 1 3i 0 ¡ ¡ 6 0 0 1 3 3i7 6 0 0 0 3 3i7 6 ¡ ¡ 7 6 ¡ ¡ 7 4 5 4 5 R3/ (1 3i) R3 ¡ ¡ R1 R4 ( 3 3i) R1 0 0 0 0 ¡ ¡ ¡ ! R4/ ( 3 3i) R4 1 3i 0 0 ¡ ¡ ¡ 2¡ ¡ 3 .

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    8 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us