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HW11solutions

# HW11solutions - Math 221 Solutions for assignment due Nov...

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Math 221, Solutions for assignment due Nov. 11 Section 7.2 Problem 10 det 3 - λ 0 4 0 - 1 - λ 0 - 2 7 3 - λ = ( - 3 - λ )( - 1 - λ )(3 - λ ) - 2( - 4( - 1 - λ )) = - ( λ +1) 2 ( λ - 1) So the matrix has eigenvalues λ = 1, with algebraic multiplicity 1 and λ = - 1, with algebraic multiplicity 2. Problem 12 det 2 - λ - 2 0 0 1 - 1 - λ 0 0 0 0 3 - λ - 4 0 0 2 - 3 - λ = det bracketleftbigg 2 - λ - 2 1 - 1 - λ bracketrightbigg det bracketleftbigg 3 - λ - 4 2 - 3 - λ bracketrightbigg = ( λ 2 - λ )( λ 2 - 1) = λ ( λ + 1)( λ - 1) 2 . So The eigenvalues are 0 , - 1 (with algebraic multiplicity 1) and - 1 (with algebraic multi- plicity 2). Problem 19 This is true. The characteristip polynomial is λ 2 - ( trace ) λ - det with det < 0, so the discriminant of this polynomial is D = trace 2 - 4 det . Since the term under the square root sign is strictly positive, the characteristic polynomial has two distinct roots: trace + D 2 and trace - D 2 . Problem 22 det( A - λI ) = det( A - λI ) T = det( A T - λI ). Thus the characteristic polynomials of A and A T are the same, which implies that the eigenvalues are the same. Problem 22 If B = S - 1 AS , then det( A - λI ) = det S - 1 ( A - λI ) S = det( S - 1 AS - S - 1 λIS ) = det( S - 1 AS - λI ) = det( B - λI ) . Thus A and B have the same characteristic polynomials, and the same eigenvalues with the same multiplicities.

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HW11solutions - Math 221 Solutions for assignment due Nov...

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