# 5.5 - Section 5.5: Complex Eigenvalues Recall: (i) λ is an...

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Unformatted text preview: Section 5.5: Complex Eigenvalues Recall: (i) λ is an ev. of A ⇔ ⇔ characteristic equation p(λ) = det(A − λI ) = 0 (A − λI )x = 0 for x = 0 (ii) by the Fundamental Thm of Algebra there are n real or complex roots {λ1, · · ·, λn } so p(λ) = a(λ − λ1 )(λ − λ2 )· · ·(λ − λn ) ∴ Every square matrix has at least one ev. Q: What does this say about rotation matrices? A: They must have complex evs. Note: Since Ax = λx, if the ev is complex (Im(λ)= 0) then so is the e-vector EX: Consider A = 0 −1 1 =⇒ A − λI = −λ −1 1 −λ 0 p(λ) = λ2 + 1 = 0 ⇒ λ1,2 = ±i, can verify have two linearly independent 1 ±i 1 i ∈ C 2. e-vectors, v2,1 = Check: Av2 = 0 −1 1 0 = −i 1 1 = −i 1 i = λ2 v2 EX: For A = for each. 43 −3 4 operating on C 2 , ﬁnd evs and eigenspace Solution: Characteristic polynomial is p(λ) = (4 − λ)(4 − λ) − (−3)(3) = λ2 − 8λ + 25 = 0 =⇒ by quadratic formula λ± = √ 8± −36 2 = 4 ± 3i −3i For λ1 = 4 + 3i : A − (4 + 3i)I = so v1 = x1 x2 3 −3 −3i , satisfying (A − (4 + 3i))v1 = 0 implies (−3i)x1 + 3x2 = 0 −3x1 − 3ix2 = 0 Note: It is tricky to use complex arithmetic to eliminate relationship, so only need to use one!! From second eqn: ∴ −3x1 = 3ix2 =⇒ x1 = −ix2 =⇒ −i 1 −i 1 unknowns, but in 2 × 2 case, both eqns give the same Basis vector for ﬁrst eigenspace is v1 = 43 −3 4 −i 1 −4i + 3 Check: A = = 3i + 4 = (4+3i) 2 Similarly, ﬁnd v2 = i 1 Note: Could instead consider e-vectors as u1 = iv1 = 1 , u2 = −iv2 = 1 −i , which is consistent i with a form we see later. parts of v , Rev and Imv , are vectors formed by real and imaginary parts of components of v ¯ For complex vectors v and matrices A, the conjugates v ¯ and A are formed by taking the conjugates of each term in the vector and the matrix. Get most properties of conjugate we had in Appendix B, e.g., ¯¯ ¯¯ rx = r x, Ax = Ax, AB = AB , rA = r A. ¯¯ ¯¯ Now ready to analyze Evs/E-vectors of Real Matrix Acting on C n : If A ∈ Rn×n has ev/e-vector pair λ ∈ C , x ∈ C n then ¯ ¯¯ Ax = Ax = λx = λx ¯¯ λ, x are also an ev/e-vector pair =⇒ Deﬁnition: For a vector v ∈ C n , the real and imaginary 3 EX above: A= 43 −3 4 has ev/e-vector pairs 1 i λ1 = 4 + 3i, u1 = and 1 −i ¯ λ2 = 4 − 3i = λ1 u2 = More generally, if v= a + bi c + di a − bi = a c ¯ = v1 +i b d = Rev + iImv , then ¯ v= c − di = Rev − iImv Imλ b (a, b) a Reλ Extension of Example above: 4 Theorem: If C = evs are λ1,2 1 . Also, if r = |λ| = |λ1 | = ±i C=r a/r −b/r b/r = r0 0r ba = a ± b, with corresponding e-vectors u2,1 = a2 + b2 , then a −b , where a, b ∈ R, then the a/r cos (φ) − sin (φ) sin (φ) cos (φ) , where the angle φ is argument of λ – that is, the angle between x−axis and line from (0, 0) to (a, b) rotation by angle φ and a scaling by r = |λ| x2 ∴ The transformation x → Cx is a composition of a multiply by A x2 Ax x φ x1 x x1 EX: A= 43 −3 4 has evs 4 ± 3i. The scale factor for the 5 From trigonometry, the angle of rotation φ is arctan (b/a) = arctan ( − 3/4) = φ (see top ﬁgure) EX: Consider the matrix A = 1 −1 . Find an invertible a −b b a such that transformation x → Ax is r = |λ1 | = 42 + (3)2 = 5. .4 .6 matrix P and matrix of form C = A = P CP −1 . corresponding to λ = .8 − 6i, consider A − (.8 − .6i)I = .2 + .6i .4 −.2 + 6i =⇒ −1 so the evs of A are λ = .8 ± .6i. To ﬁnd an e-vector Solution: The characteristic polynomial is λ2 − 1.6λ +1, . (A − (.8 − .6i)I )x = 0 x2 free. .4x1 + (−.2 + .6i)x2 = 0 =⇒ x1 = ((1 − 3i)/2)x2 with 1 − 3i 2 A corresponding e-vector is v = . Choosing (for reasons we see shortly) P = [Rev Imx] = 1 −3 2 =⇒ P −1 = 0 1/2 −1/3 1/6 0 6 C = P −1 AP = 1 6 03 −2 1 .4 1 −1 .6 1 −3 2 = 0 .6 .8 −.6 .8 ∴ See that P gives a change of variables x = P y followed by a rotation, and then a return to the orginal variable. Can show this gives a rotation acting like an ellipse instead of a circle (see ﬁgure in book). In general, can show λ = a − bi (b = 0) and corresponding e-vector v ∈ C 2 , a −b P = [Rev Imx] =⇒ A = P CP −1 , for C = ba A Theorem: For any 2 × 2 matrix A with complex ev x Ax Change of variable P −1 Change of variable P u C rotation Cu 7 ...
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## This note was uploaded on 04/09/2011 for the course MATH 232 taught by Professor Russel during the Spring '10 term at Simon Fraser.

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