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102_solutions8

# 102_solutions8 - MATH 102 SOLUTIONS TO HW#8 Section 4.4...

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Unformatted text preview: MATH 102 SOLUTIONS TO HW #8 Section 4.4, problem 6. A simple example of this is the matrix: A = 1 . A quick calculation shows that p A ( Î» ) = | Î»I- A | = Î» 2 , so that Î» = 0 is the only eigenvalue of A . However, if one performs the row operation { Row 1 } + { Row 2 } â‡’ { Row 2 } then we are left with the matrix: B = 1 1 . We have that p B ( Î» ) = Î» ( Î»- 1), so that B has eigenvalues Î» = 0 , 1. Notice that the eigenvalue Î» = 0 persisted in the above row operation (i.e. from A to B ). This is because the eigenvectors of Î» = 0 are clearly the null-space of A or B . And the null-space of a matrix is not changed by row operations. Section 5.1, problem 10. a) Here is a simple example of this. Consider the matrices: A = 1 1 , B = 1- 1 . These have the characteristic polynomials p A ( Î» ) = Î» 2- 1 and p B ( Î» ) = Î» 2 + 1. Thus A has eigenvalues Î» Â± = Â± 1, and B has eigenvalues Î» Â± = Â± i ....
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102_solutions8 - MATH 102 SOLUTIONS TO HW#8 Section 4.4...

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