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Unformatted text preview: Supplementary Assignment Math 417 — Winter 2009 Do Not Submit § 8.1#8. We first find the eigenvalues of A by computing its characteristic polynomial det( A λI 2 ) = det 3 λ 3 3 5 λ = (3 λ )( 5 λ ) 9 = ( 6 λ )(4 λ ) . Then we compute the corresponding eigenspaces. λ = 6: E 6 = ker 9 3 3 1 = ker 1 1 / 3 = span 1 / 3 1 . λ = 4: E 4 = ker 1 3 3 9 = ker 1 3 = span 3 1 . The two eigenvectors we found are necessarily orthogonal to each other since they have different eigenvalues. Therefore, we can find an eigenbasis for A by normalizing each one: 1 √ 10 1 3 , 1 √ 10 3 1 . Using these two vectors, we find the orthogonal matrix S = 1 √ 10 1 3 3 1 Since S is othogonal, the inverse of S is simply S T and 6 0 4 = S T AS. § 8.1#8*. For this problem, we are asked to represent the matrix A as a linear combination of projection matrices as in Version IV of the Spectral Theorem discussed in class. From the above the eigenvalues of A are 6 and 4, so we can write A = ( 6) P 6 + (4) P 4 1 where P 6 is the matrix of the orthogonal projection onto the eigenspace E 6 and P 4 is the matrix of the orthogonal projection onto the eigenspace E 4 . From the previous solution, the unit vector 1 / √ 10 3 / √ 10 is by itself an orthonormal basis for E 6 . Therefore, P 6 = 1 / √ 10 3 / √ 10 1 / √ 10 3 / √ 10 = 1 10 1 3 3 9 . Similarly, from the previous solution, the unit vector 3 / √ 10 1 / √ 10 is by itself an orthonormal basis for E 4 . Therefore, P 4 = 3 / √ 10 1 / √ 10 3 / √ 10 1 / √ 10 = 1 10 9 3 3 1 . Finally, we can check that A = 3 3 3 5 = 6 10 1 3 3 9 + 4 10 9 3 3 1 = ( 6) P 6 + (4) P 4 . § 8.1#10. As usual, we first find the eigenvalues of A . We can evaluate the characteristic polynomial by cofactor expansion along the first row: det( A λI 3 ) = det 1 λ 2 2 2 4...
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This note was uploaded on 10/13/2009 for the course MATH 417 taught by Professor Eveet during the Spring '09 term at University of MichiganDearborn.
 Spring '09
 eveet
 Math, Linear Algebra, Algebra

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