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Unformatted text preview: Solutions to 7.4: 2, 3, 4, 7, 8 2. Note that det( A λI ) = 0 ⇐⇒ λ 4 4 λ = 0 ⇐⇒ λ 2 + 16 = 0 ⇐⇒ λ = 4 i or λ = 4 i. Eigenvectors for λ = 4 i : Here we have A 4 iI = 4 i 4 4 4 i , and so we have an eigenvector corresponding to nullspace( A 4 iI ) as follows: v = 1 i . Hence, we obtain the solution x ( t ) = e 4 it 1 i = (cos4 t + i sin4 t ) 1 + i 1 = cos4 t 1 sin4 t 1 + i cos4 t 1 + sin4 t 1 . Therefore, we obtain the two realvalued solutions x 1 ( t ) = cos4 t 1 sin4 t 1 and x 2 ( t ) = cos4 t 1 + sin4 t 1 . Putting the two solutions x 1 ( t ) and x 2 ( t ) together, we obtain the general solution to this linear system: x ( t ) = c 1 cos4 t 1 sin4 t 1 + c 2 cos4 t 1 + sin4 t 1 . Equivalently, this can be expressed as x ( t ) = c 1 cos4 t sin4 t + c 2 sin4 t cos4 t . Note that one can also obtain the solution by working with the eigenvalue λ = 4 i . The calculations are similar. 1 3. Note that det( A λI ) = 0 ⇐⇒ 1 λ 2 5 5 λ ⇐⇒ λ 2 + 4 λ + 5 = 0 ⇐⇒ λ = 2 ± i. Eigenvectors for λ = 2 + i : Here we have A ( 2 + i ) I = 3 i 2 5 3 i . The two rows of this matrix must be proportional, since we must be able to find a nonzero vector in its nullspace corresponding to λ = 2 + i , so concentrating on the first row, we can determine an eigenvector corresponding to λ = 2 + i as follows: v = 2 3 i . Note that other choices of v are possible here, such as 3 + i 5 , obtained by concentrating on the second row instead of the first row. Proceeding with our choice of v , we obtain the solution x ( t ) = e ( 2+ i ) t 2 3 i = e 2 t (cos t + i sin t ) 2 3 + i 1 = e 2 t cos t 2 3 sin t 1 + i cos t 1 + sin t 2 3 . Therefore, we obtain the two realvalued solutions x 1 ( t ) = e 2 t cos t 2 3 sin t 1 and x 2 ( t ) = e 2 t cos t 1 + sin t 2 3 . Putting the two solutions x 1 ( t ) and x 2 ( t ) together, we obtain the general solution to this linear system: x ( t ) = c 1 e 2 t cos t 2 3 sin t 1 + c 2 e 2 t cos t 1 + sin t 2 3 . Equivalently, this can be expressed as x ( t ) = c 1 e 2 t 2cos t 3cos t + sin t + c 2 e 2 t 2sin t cos t + 3sin t . Note that one can also obtain the solution by working with the eigenvalue λ = 2 i . The calculations are similar. 2 4. Note that det( A λI ) = 0 ⇐⇒ 1 λ 2 2 1 λ = 0 ⇐⇒ λ 2 + 2 λ + 5 = 0 ⇐⇒ λ = 1 ± 2 i. Eigenvectors for λ = 1 + 2 i : Here we have A ( 1 + 2 i ) I = 2 i 2 2 2 i . One choice of eigenvector, found by examining the nullspace of this matrix, is v = 1 i . We obtain the solution x ( t ) = e ( 1+2 i ) t 1 i = e t (cos2 t + i sin2 t ) 1 + i 1 = e t cos2 t 1 sin2 t 1 + i cos2 t 1 + sin2 t 1 . Therefore, we obtain the two realvalued solutions x 1 ( t ) = e t cos2 t 1 sin2 t 1 and x 2 ( t ) = e t cos2 t 1 + sin2 t 1 ....
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 Fall '10
 Roberts
 Cos, Eigenvalue, eigenvector and eigenspace, general solution

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