Linear Algebra with Applications (3rd Edition)

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Math 417 homework 8 solutions Problem 1 The matrix is cos α - sin α sin α cos α . Characteristic polynomial: (cos α - z ) 2 + sin 2 α Roots: solve (cos α - λ ) 2 + sin 2 α = 0 . Get: (cos α - λ ) 2 = - sin 2 α, so cos α - λ = ± i sin α, so λ = cos α ± i sin α. Problem 2 Result: eigenvalues - 1 (algebraic multiplicity 2) and 2 (algebraic multiplicity 1). For the first eigenvalue, we obtain an eigenvector 0 1 0 . The dimension of this eigenspace, i.e. the geometric multiplicity of - 1, is 1, less than the algebraic multiplicity. For the other eigenvalue, an eigenvector is 1 0 - 1 . Again, geometric multiplicity 1. In this case the matrix is not diagonalizable because we cannot find a third eigenvector that is not redundant. 1
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Problem 3 First write the right-hand side in polar form: w = - 16 = 16 · e = re with r = 16, φ = π . Now use the formula from class: the n solutions are z = r 1 /n exp i ( φ + 2 πk ) n where k = 0 , . . . , n - 1 (with n = 8). So: k z 0 2 e 8 1 2 e 3 8 2 2 e 5 8 3 2 e 7 8 4 2 e 9 8 5
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