Unformatted text preview: y = e Î»t , we get from y 00 = 5 y + 4 y that Î» 2 = 5 Î» + 4 giving the same roots. 17 ) (a) We need a matrix with both eigenvalues real, one positive an one negative. Â±1 0 0 1 Â² works. (b) We need a matrix with both eigenvalues real and both positive. Â± 1 0 0 1 Â² works. (c) We need a matrix with both eigenvalues complex and real part positive. Â± a bb a Â² works, where a > 0. 19 ) e Bt = I + âˆ‘ âˆž k =1 t k B k /k ! = I + tB + + + ... = I + tB = Â± 14 t 1 Â² . Itâ€™s derivative is Â±4 Â² while Be Bt = Â±4 Â²Â± 14 t 1 Â² = Â±4 Â² . 25 ) Â± 1 3 0 0 Â² 2 = Â± 1 3 0 0 Â² . Note A 3 = A 2 A = AA = A . Similarly, A 4 = A 3 A = AA = A . Continuing on, A s = A s1 A = AA = A . (This is called proof by induction ). So e At = I + âˆ‘ âˆž k =1 t k A k /k ! = I + âˆ‘ âˆž k =1 t k A/k ! = I + ( e t1) A = Â± e t 3( e t1) Â² ....
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 '05
 HUI
 Math, Linear Algebra, Algebra, Characteristic polynomial, Complex number

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