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Course 111: Algebra, 20th April 2007 To be handed in at tutorials on April 23rd and 24th. 1. Consider the matrix 1 0 1 A = 2 −2 −1 0 0 1 Write A in its Jordan form, J by determining the matrix V which transforms A according to J = V −1 AV . A has characteristic equation (1 − λ)2 (−2 − λ) = 0 and so λ(A) = {1, 1, −2} are the eigenvalues. When λ = −2 the corresponding eigenvector is v1 = (0, 1, 0) When λ = 1 there is one corresponding eigenvector, v2 = (3/2, 1, 0). Solving Av3 = v2 for the generalised eigenvector v3 gives (0, 0, 0) (not useful). Instead consider the related expression (A − λI)2 v3 = 0 (as observed in the notes). For λ = 1 this gives 0 0 0 x −6 9 5 y = (0, 0, 0) 0 0 0 z yielding −6x + 9y + 5z = 0 and any (x, y, z) satisfying this expression is a generalised eigenvector. Thus, v3 = (−11/6, 2/3, 1) is appropriate. Then 0 3/2 −11/6 V = 1 2/3 1 0 0 1 and ⇒ V −1 − 23 1 − 17 9 11 = 23 0 9 0 0 1 and therefore, J = V −1 AV − 23 1 − 17 1 0 1 0 3/2 −11/6 9 11 2 −2 −1 1 2/3 = 23 0 1 9 0 0 1 0 0 1 0 0 1 −2 0 0 = 0 1 1 0 0 1 Use the formula in your notes to determine e2J . Writing J in its block diagonal form J= and so e 2J = J1 0 0 J2 ! e2J1 0 2J2 0 e ! where e2J1 = (e2(−2) ) = (e−4 ) and using the formula in the notes e 2J2 =e 2(1) 1 t=2 0 1 ! =e 2(1) 1 2 0 1 ! and so e2J e−4 0 0 0.0183 0 0 2 2 0 7.389 14.778 = 0 e 2e = 0 0 e2 0 0 7.389 2. Consider the matrix equation, Ax = b with A= 1 1 1 1+ ! , 0 < << 1 (1) Determine the condition number of the matrix A with respect to the ∞ norm. ||A||∞ = max{2, 2 + } = 2 + and ||A−1 ||∞ = max{(2 + ), 2/} = (2 + )/. Then κ∞ = ||A||∞||A−1 ||∞ = (2 + )(2 + )/. Is the matrix well or poorly conditioned? The matrix is poorly conditioned since for 0 < << 1 the condition number is large. Given 2 2 b= ! , determine a solution, x to the matrix equation Ax = b. The solution is x = (2, 0)T . Determine what happens to the solution when A= 1 1 1 1 ! , 0 < << 1 and b is defined as above. The solution is now any x = (x1 , x2 ) with x1 + x2 = 2. Also, determine the effect of a small change in b on the solution by considering solutions when b = (2, 2)T and b = (2 + , 2)T and A is as given in Eqn 1. For b = (2, 2)T the solution is x = (2, 0)T . For b = (2 + , 2)T the solution is x = (1, 1)T .