Introduction to the square root of a 2 by 2 matrix Introduction to the square root of a 2 by 2 matrix Introduction to the square root of a 2 by 2 matrix Introduction to the square root of a 2 by 2 - PDF Document

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  1. Introduction to the square root of a 2 by 2 matrix Introduction to the square root of a 2 by 2 matrix Introduction to the square root of a 2 by 2 matrix Introduction to the square root of a 2 by 2 matrix Yue Kwok Choy Yue Kwok Choy Yue Kwok Choy Yue Kwok Choy The square root of a 2 by 2 matrix A A A A is another 2 by 2 matrix B B B B such that ? = ??, where ?? ? ?= ?. In general, there can be no, two, four stands for the matrix product of B B B B with itself. We write ? or even an infinite number of square root matrices. For example: ?1 2 4??1 2 4? = ?7 10 22? For example: For example: For example: 3 3 15 ? ?= ?1 If we take ? = ?7 10 22? , then ? 4? is one of the square roots of ? = ?7 ? ?= ?7 10 22? 2 4? 15 15 3 Note that ? = ?1 2 10 22? . 3 15 Questions Questions Questions Questions (See answers at the end of this file) (See answers at the end of this file) (See answers at the end of this file) (See answers at the end of this file) There may be another root(s) for ?. Evaluate : ?1 1 1??1 1 1?. Hence find two (a) ?2 2 2? (b) (b) ?1 1 1? . 1. 1. 1. 1. two two two square roots of (a) (a) (a) (b) (b) 1 1 2 1 ? ? . Hence find four ? ? , where a,b ≥ 0 . four values of ?4 0 9? four values of ?a 0 b? 2. 2. 2. 2. Find four four four four four 0 0 List all possible matrices ? = ?a b d? , where a, b, c and d can be either 0 or 1, such that ? ? ? 3. 3. 3. 3. c (A) (A) (A) (A) Solving equations Solving equations Solving equations Solving equations method does not exist. method method method Suppose we like to find the square root of ? = ?1 2 5?. So we write: 2 2 4? ⟹ ?a?+ bc c(a + d) b(a + d) cb + d?? = ?1 2 5? ?a b d??a b d? = ?1 2 2 c c We get four equations: a?+ bc = 1 …(1) b(a + d) = 2 …(2) c(a + d) = 2 …(3) cb + d?= 5 …(4) (2) − (3), (b − c)(a + d) = 0 Since a + d ≠ 0 (otherwise (2) or (3) give absurdity), b − c = 0, b = c. Hence we get: a?+ b?= 1 …(5) b(a + d) = 2 …(6) 1

  2. b?+ d?= 5 …(7) ? (6)?, b?(a + d)?= 4 , b?= (???)? …(8) ? (8) ↓ (5), a?+ (???)?= 1, …(9) (7) − (5),d?− a?= 4 …(10) ? d − a = ??? …(11) (a + d ≠ 0) (11) ↓ (9), a?+??? ???= 1, a?+ a?d + d − a = a + d, a?+ a?d − 2a = 0 a(a?+ ad − 2) = 0 ???? ? …(12) ∴ a = 0 or d = ? ? (12) ↓ (10), a = 0,d = ±2 or a = ± √?,d = ± √? The rest can be solved easily, we therefore have: ? ? ± ± ? ?= ?0 ±1 ±2?, ? ?1 2 2 5? √? ? √? √? ? √? ? ±1 ± ± ? ? does not exist. Prove that ?0 1 0? 3. 3. 3. 3. 0 List, without prove, all possible matrices ? = ?a b d? , where a, b, c and d can be either 0 or 1, c ? ? has no real solution(s). such that ? ? ?. Use algebraic method to find ?1 2 1? 4. 4. 4. 4. 0 5. 5. 5. 5. Check that the square root of the identity matrix is given by: ? ?= ?±d ±d ???? ? c ???? ? ∓d ?1 0 0 1? ?,? ∓d? , c where ±?±1 ∓1?,?±1 0 ∓1?,?±1 0 c ∓1? are limiting cases. 0 c 0 (a) Let ? = ?1 2 4?, find |?| and ??. Hence find ? ? ? . 6. 6. 6. 6. (a) (a) (a) 2 (b) ? = ?a b d?, where |?| = 0. Given that tr(?) = a + d > 0. (b) (b) (b) c ? ? . Show that ??= tr(?)? . Hence find ? 2

  3. (B) (B) (B) (B) Diagonalization of Matrrix Diagonalization of Matrrix Diagonalization of Matrrix Diagonalization of Matrrix Solving equation method in finding the square root of a matrix may not be easy. It involves ? ?, and soon solving four non-linear equations with four unknowns. You may try this: ?41 12 34? 12 may give up. We note that the square root of a diagonal ? ?= ?√a 0 If a matrix is NOT a diagonal matrix, we devise a method called diagonalization to help us. diagonal diagonal diagonal matrix can be found easily: √b?,?−√a 0 √b?,?√a 0 −√b?,?−√a 0 0 ?a 0 0 b? −√b?. 0 0 0 We proceed with the finding of the eigenvalue(s) and eigenvector(s) of A . A real number λ is said to be an eigenvalue of a matrix A if there exists a non-zero column vector v such that Av = λv or (A − λI)v = 0 . ? ? We like to find ?33 24 57? 48 ( ( ( (1 1 1 1) ) ) ) Eigenvalues Eigenvalues Eigenvalues Eigenvalues 24 57?, v = ?x 57 − λ??x A = ?33 y?, (A − λI)v = ?33 − λ 24 y? = 0 48 48 Now, (A − λI)v = 0 has non-zero solution, |A − λI| = 0 ?33 − λ 48 (33 − λ)(57 − λ) − 48 × 24 = 0 λ?− 90λ + 729 = 0 (λ − 81) (λ − 9) = 0 ∴ λ = 9 or λ = 81 , and these are the eigenvalues ( ( ( (2 2 2 2) ) ) ) Eigenvectors Eigenvectors Eigenvectors Eigenvectors We usually would like to find the eigenvector corresponding to each eigenvalue. The process is called normalization. For λ?= 9, ?33 48 Choose for convenience 24x?+ 24y?= 0 ⇔ x?+ y?= 0 ∴?x? −1?, which is a eigenvector For λ?= 81, ?33 48 24 57 − λ? = 0 eigenvalues eigenvalues eigenvalues. 24 57??x? y?? = 9?x? ?33x?+ 24y?= 9x? 48x?+ 57y?= 9y? y?? ? y?? = ?1 eigenvector eigenvector eigenvector. 24 57??x? y?? = 81?x? y??⇔?33x?+ 24y?= 81x? 48x?+ 57y?= 81y? ? 3

  4. Choose −48x?+ 24y?= 0 ⇔2x?− y?= 0 y?? = ?1 ∴?x? 2?, which is another eigenvector. ( ( ( (3 3 3 3) ) ) ) Diagonalization of matrix Diagonalization of matrix Diagonalization of matrix Diagonalization of matrix We place two eigenvectors together. Let ? = ?x? x? y?? = ?1 1 2? y? −1 Consider another matrix: ?? ?33 48 24 57??1 ? = ????? = ?1 1 2? 1 2? −1 −1 −1 1??33 24 57??1 ? ??2 1 2? = 48 1 −1 = ?9 0 81? 0 At last we get a matrix B B B B which is diagonal with eigenvalues as entries in the main diagonal. ? ?= ?3 ? ?= ?9 0 0 9?,?−3 −9?,?3 0 −9?,?−3 0 0 9? ? 81? 0 0 0 0 0 Note: Note: Note: Note: may have equal roots or even complex roots. Although most matrices are not diagonal, they can be diagonalized. Not all square matrices can be diagonalised.A thorough study of diagonalization of a matrix is not discussed here. ( ( ( (4 4 4 4) ) ) ) Find Find Find Finding the ing the ing the ing the square root square root square root square root of the original matrix A of the original matrix A of the original matrix A of the original matrix A Since ? = ?????, we have ? = ????? The diagonalization of a matrix may not be a simple subject since |A − λI| = 0 ? ???????? ? ????? = ?? ? ?(????)? ? ????= ?? ? ?? ? ????= ?????= ? ??? ? ? ????? So ??? = ? ?? 1 2??3 0 9??1 1 2??3 0 9??? ? ????= ?1 1 2? = ?1 ??2 −1 1?? ? ?= ?? (a) ? 0 0 −1 −1 −1 1 = ?3 9 18?? 3 6??2 2 7? ??2 −1 1? = ?1 −1 1? = ?5 −3 1 −1 1 4 ?? 1 2??−3 0 9??1 1 2??−3 0 9??? ? ?P??= ?1 1 2? = ?1 ??2 −1 1?? ? ?= PB (b) A 0 0 −1 −1 −1 1 = ?−3 9 18?? 3 6??2 −1 1? = ?1 4 5? ??2 −1 1? = ?−1 3 8 1 1 1 ? ?= ?? ∴ ?33 24 57? −? −??,?? ? ??,?−? −? −?? ? ??,?−? 48 ? −? ? −? 4

  5. Questions Questions Questions Questions 7. 7. 7. 7. Real matrix may have irrational ? ?= ?1 irrational irrational irrational square roots. Check by multipication multipication multipication multipication: ?1 0 2 3? √3 − 1 √3 ?,?1 −√3 − 1 −√3 ?,?−1 √3 + 1 √3 ?,?−1 −√3 + 1 −√3 ? 0 0 0 0 A simple r r r real 8. 8. 8. 8. eal eal eal matrix may have complex complex complex complex square roots. Check by multipication multipication multipication multipication: ? ?= ??1 + i 1 − i 1 − i 1 + i?,? ??1 − i 1 + i 1 + i 1 − i?,? ??−1 − i −1 + i −1 + i −1 − i?,? ??−1 + i −1 − i −1 − i −1 + i? ?0 1 1 0? ? 9. 9. 9. 9. A matrix can have both integral and fractional square roots. Use Diagonalization Method to show: ? ? ? ? ? ? ? ?= ±?2 ?0 4 5? −4 −3?,±? ? −? −1 1 ? Answers Answers Answers Answers ?1 1 1 1??1 1 1? = ?2 2 2? , Hence: 1. 1. 1. 1. 1 2 ? ?= ?1 (a) ?2 2 2? 1 1??1 1 1?, ?−1 1? = ?2 −1 −1? (a) (a) (a) 2 1 1 −1 2 2? = 2?1 (b) ?1 1 1? ⟹ ?1 1 1? = ?± √??1 1 1???± √??1 1 1?? ? ? (b) (b) (b) 1 1 2 1 1 1 1 ? ? ? ? − − ? ?= ± ?1 1 1 1? √??1 1 1? = ? ? √? ? √? √? ? √? √? ? √? √? ? √? ?, ? ? 1 − − ? ?= ?2 ?4 0 0 9? 0 3?,?−2 0 3?,?2 −3?,?−2 0 0 2. 2. 2. 2. −3? 0 0 0 0 ? ?= ?√a √b?,?−√a 0 √b?,?√a 0 −√b?,?−√a 0 0 ?a 0 0 b? −√b? 0 0 0 0 1 0? ⟹ ?a?+ bc c(a + d) b(a + d) cb + d?? = ?0 b d? = ?0 1 0? ?a b d??a 3. 3. 3. 3. 0 0 c c a?+ bc = 0 = cb + d?⟹ a = ±d (a) Since b(a + d) = 1,a + d ≠ 0, and so a = d ≠ 0. (b) Lastly, since c(a + d) = 0 ⟹ c = 0 and so a?+ bc = a?= 0, contradicts with (a). ? ?,?0 ? ? does not exist ?0 0 1 0? 0 0? 1 ? ??1 ? ? has no real solution(s). 1 0? ?0 1 ?a c 1 1? b d??a 1 2 1? ⟹ ?a?+ bc c(a + d) b(a + d) cb + d?? = ?1 b d? = ?1 2 1? 4. 4. 4. 4. 0 0 c 5

  6. We get four equations: a?+ bc = 1 …(1) b(a + d) = 2 …(2) c(a + d) = 0 …(3) cb + d?= 1 …(4) From (3), c = 0 …(5) (a + d = 0 gives contradiction in (2)) From (1), a?= 1, a = ±1. From (3), d?= 1, d = ±1 ? ?= ?1 Hence, ?1 2 1? 2 4?, |?| = 0 , ??= ?1 1 1?,?−1 −1 −1? 0 0 0 (a) ? = ?1 2 4??1 2 4? = ?5 10 20? = 5?1 2 4? = 5?. 6. 6. 6. 6. (a) (a) (a) 2 2 2 2 10 ? ? ? . ? √??1 2 4? = ? ? ?= ? ???= ?? ? ? √? ? √? √? ? √? Hence ? = √??? √?? = ? . 2 (b) ? = ?a b d?, |?| = ad − bc = 0 ⟹ ad = bc …(1) d? = ?a?+ bc c(a + d) = ?a(a + d) c(a + d) (b) (b) (b) c b(a + d) cb + d?? = ?a?+ ad b(a + d) ad + d?? , by (1). ??= ?a b d??a b c c c(a + d) b(a + d) d(a + d)? = (a + d)?a b d? = tr(?)? c ? ? ? ??(?)??= ? , where tr(?) > 0. ∴ ? = ???(?)?? ? ?= ? Hence, ? ???(?)? . 9. Eigenvalues : 4, 1 1?,?4 Eigenvectors: ?1 1? ? = ?x? x? y?? = ?1 4 1?,???= −1? , ? = ?4 4 0 1? ? ??−1 y? 0 1 1 ? ?= ?2 ? ?= ?4 0 1? 0 1?,?−2 −1?,?2 0 −1?,?−2 0 0 1? ? 0 0 0 0 0 ? ? ? ? ? ? ?? 4 1??2 0 1??1 4 1? 4 1??2 0 1??? 4 ? ????= ?1 = ?1 ??−1 ? ?= ?? (a) (a) (a) (a) ? −1?? = ? ? −? 0 0 1 1 1 1 ? ?? 4 1??−2 0 1??1 4 1? 4 1??−2 0 1??? −1?? = ?2 4 −4 −3? ? ????= ?1 = ?1 ??−1 ? ?= ?? (b) (b) (b) (b) ? 0 0 1 1 1 1 1 ? ? ? ? ? ? ? ?= ±?2 ∴ ?0 4 5? −4 −3?,±? ? −? −1 1 ? 6