Linear Algebra Solutions 68

Linear Algebra Solutions 68 - a d Î ad-bc has a repeated...

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Hence z = z (0) e ( a - ib ) t x 1 + iy 1 = ( x 1 (0) + iy 1 (0)) e at (cos bt - i sin bt ) . Equating real and imaginary parts gives x 1 = e at { x 1 (0)cos bt + y 1 (0)sin bt } y 1 = e at { y 1 (0)cos bt - x 1 (0)sin bt } . Now if we deFne α and β by α β = P - 1 x (0) y (0) , we see that α = x 1 (0) and β = y 1 (0). Then x y = P x 1 y 1 = [ U | V ] e at ( α cos bt + β sin bt ) e at ( β cos bt - α sin bt ) = e at { ( α cos bt + β sin bt ) U + ( β cos bt - α sin bt ) V } = e at { cos bt ( αU + βV ) + sin bt ( βU - αV ) } . 7. (The case of repeated eigenvalues.) Let A = a b c d and suppose that the characteristic polynomial of A , λ 2
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Unformatted text preview: -( a + d ) Î» +( ad-bc ), has a repeated root Î± . Also assume that A 6 = Î±I 2 . (i) Î» 2-( a + d ) Î» + ( ad-bc ) = ( Î»-Î± ) 2 = Î» 2-2 Î±Î» + Î± 2 . Hence a + d = 2 Î± and ad-bc = Î± 2 and ( a + d ) 2 = 4( ad-bc ) , a 2 + 2 ad + d 2 = 4 ad-4 bc, a 2-2 ad + d 2 + 4 bc = , ( a-d ) 2 + 4 bc = . 71...
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