Linear Algebra with Applications (3rd Edition)

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Math 417 homework 9 solutions Problem 1 Characteristic polynomial: χ A ( z ) = det( A - zI ) = det - 3 - z - 2 0 2 - 3 - z 0 0 0 4 - z = ( ( - 3 - z ) 2 - 2( - 2) ) (4 - z ) = ((3 + z ) 2 + 4)(4 - z ) . One root: z = 4. The other two are obtained by solving 0 = (3 + z ) 2 + 4 which yields z = - 3 ± 2 i. The matrix is simple: there are three different eigenvalues, each has algebraic multiplicity 1. Geometric multiplicity is always algebraic, so it must be 1 as well. Hence each eigenspace has dimension 1; it is sufficient to find one eigenvector. Eigenvector for first eigenvalue: A - 4 I = - 3 - 2 0 2 - 3 0 0 0 0 . The kernel is spanned by ~v 1 = 0 0 1 . Eigenvector for second eigenvalue: A - ( - 3 + 2 i ) I = - 2 i - 2 0 2 - 2 i 0 0 0 7 - 2 i . One kernel vector: ~v 2 = 1 - i 0 . 1
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Instead of calculating the third kernel vector, we use from class: Fact: if A is real and A~v = λ~v , then A ~v = λ ~v . The third eigenvalue is the conjugate of the second, so the third eigenvector is a conjugate of the second (or any nonzero multiple of that).
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