MATH311ansHW9

# MATH311ansHW9 - Math 311 Homework 9 partial answers and...

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Unformatted text preview: Math 311, Homework 9 partial answers and solutions 3.6C.2. Answer: Theorem 6.7 does not apply, no basis of eigenvectors. 3.6C.3. The characteristic polynomial det- λ 1- 1- λ = λ 2 + 1 has two distinct complex roots λ = ± i . Thus the matrix has a basis of eigenvectors in C 2 , but not in R 2 . 3.6C.8. The characteristic polynomial is det cos θ- λ- sin θ sin θ cos θ- λ = (cos θ- λ ) 2 + sin 2 θ = cos 2 θ- 2 cos θλ + λ 2 + sin 2 θ = λ 2- 2 cos θλ + 1 . Using the quadratic formula, its discriminant is D = 4(cos 2 θ- 1) =- 4 sin 2 θ, and so λ = cos θ ± i sin θ. The corresponding eigenvectors are in the null spaces of ∓ i sin θ- sin θ sin θ ∓ i sin θ , so they are ( ± i 1 ) . 3.6C.11. The characteristic polynomial is det 4- λ- 3 2- 1- λ =- 4 + λ- 4 λ + λ 2 + 6 = λ 2- 3 λ + 2 = ( λ- 1)( λ- 2) . Therefore the eigenvalues are λ = 1 , 2 . The eigenvectors corresponding to λ = 1 lie in the null space of 3- 3 2- 2 so one of them is ( 1 1 ) . The eigenvectors corresponding to....
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## This note was uploaded on 04/29/2008 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.

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MATH311ansHW9 - Math 311 Homework 9 partial answers and...

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