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Course 111: Algebra, 13th April 2007 To be handed in at tutorials on April 16th and 17th. 1. Consider the matrix 0 1 −2 −3 A= ! . Find the matrix eigenvalues and eigenvectors. Solution The characteristic equation is λ2 + 3λ + 2 = 0 with solution λ1 = −1 and λ2 = −2. For λ = −1, the corresponding eigenvector is v1 = k1 (1, −1)T for any scalar k1 . For λ = −2, the corresponding eigenvector is v2 = k2 (1, −2)T for arbitrary scalar k2 . 2. Consider the matrix 4 1 −1 6 A= ! . Show by determining the transformation matrix V such that AV = V J that the Jordan form of the matrix A is J= 5 1 0 5 ! The characteristic equation is (λ−5)2 = 0 so the eigenvalues are λ(A) = {5, 5}. The corresponding eigenvector is v1 = (1, 1)T and 1 generalised eigenvector, v2 , is needed. To get this solve (A − λI)v2 = v1 to get (0, 1). The transformation matrix is then V = 1 0 1 1 ! ⇒ V −1 = and J =V as required. −1 AV = 5 1 0 5 1 0 −1 1 ! !