M115A-AssignmentSevenExtraSolutions - MATH 115A Assignment Seven Extra Solutions Paul Skoufranis For Question 3 of Section 5.1 for each matrix A Mnn(F

M115A-AssignmentSevenExtraSolutions - MATH 115A Assignment...

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MATH 115A - Assignment Seven Extra Solutions Paul Skoufranis February 26, 2012 For Question 3 of Section 5.1, for each matrix A ∈ M n × n ( F ) , determine all eigenvalues of A and for each eigenvalue λ of A , find the set of eigenvectors corresponding to λ . If possible, find a basis of F n consisting of eigenvectors of A and determine an invertible matrix Q and a diagonal matrix D such that Q - 1 AQ = D . § 5.1 Question 3a) A = 1 2 3 2 for F = R . Solution : We desire to find the eigenvalues and eigenvector of A . First we compute the eigenvalues. To find the eigenvalues, we compute the characteristic polynomial: χ A ( λ ) = det( λI - A ) = det λ - 1 - 2 - 3 λ - 2 = ( λ - 1)( λ - 2) - 6 = λ 2 - 3 λ - 4 = ( λ - 4)( λ + 1) Thus the eigenvalues of A are 4 and - 1. Now we shall compute the eigenspaces. We notice that E 4 = ker (4 I - A ) = ker 3 - 2 - 3 2 = ker 3 - 2 0 0 = span ( { (2 , 3) } ) where the third equality comes from a simple row reduction. Therefore { (2 , 3) } is a basis for E 4 . Next we notice that E - 1 = ker ( - I - A ) = ker - 2 - 2 - 3 - 3 = ker 1 1 0 0 = span ( { (1 , - 1) } ) where the third equality comes from a simple row reduction. Therefore { (1 , - 1) } is a basis for E - 1 . Combining these two bases, we see that β = { (2 , 3) , (1 , - 1) } is an eigenbasis for A (and thus a basis for R 2 consisting of eigenvectors). Finally, to find the desired matrices Q and D , we notice that, if γ is the standard basis for R 2 , then Q - 1 AQ = D where D = 4 0 0 - 1 and Q is the change of basis matrix that takes β -coordinates to γ -coordinates. Therefore, by using the definition of the change of basis matrix Q = 2 1 3 - 1 which completes the problem.
§ 5.1 Question 3b) A = 0 - 2 - 3 - 1 1 - 1 2 2 5 for F = R . Solution : We desire to find the eigenvalues and eigenvector of A . First we compute the eigenvalues. To find the eigenvalues, we compute the characteristic polynomial: χ A ( λ ) = det λ 2 3 1 λ - 1 1 - 2 - 2 λ - 5 = λ ( λ - 1)( λ - 5) + (2)(1)( - 2) + (3)(1)( - 2) - (3)( λ - 1)( - 2) - (2)(1)( λ - 5) - λ (1)( - 2) = ( λ 3 - 6 λ 2 + 5 λ ) - 4 - 6 + (6 λ - 6) - (2 λ - 10) + 2 λ = λ 3 - 6 λ 2 + 11 λ - 6 Therefore, to find the eigenvalues of A , we need to factor χ A . One way is to notice that χ A (1) = 0 (we would try 1 as 1 divides - 6 (see the Rational Roots Theorem)) and then use long division of polynomials. Another way is to notice that χ A ( λ ) = ( λ 3 - 6 λ 2 + 5 λ ) + (6 λ - 6) = λ ( λ - 5)( λ - 1) + 6( λ - 1) = ( λ - 1)( λ ( λ - 5) + 6) = ( λ - 1)( λ 2 - 5 λ + 6) = ( λ - 1)( λ - 2)( λ - 3) Thus the eigenvalues of A are 1, 2, and 3. Now we shall compute the eigenspaces. We notice that E 3 = ker 3 2 3 1 2 1 - 2 - 2 - 2 = ker 1 0 1 0 1 0 0 0 0 = span ( { (1 , 0 , - 1) } ) where the second equality comes from a simple row reduction. Therefore { (1 , 0 , - 1) } is a basis for E 3 . Next we notice that E 2 = ker 2 2 3 1 1 1 - 2 - 2 - 3 = ker 1 1 0 0 0 1 0 0 0 = span ( { (1 , - 1 , 0) } ) where the second equality comes from a simple row reduction. Therefore { (1 , - 1 , 0) } is a basis for E 2 .

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