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Math 2270, Fall 2015 Instructor: Thomas Goller 10 November 2015 Quiz #7 Solutions (1) Let A = 1 4 2 . 6 (a) Compute the eigenvalues 1, 2 of A. (2 points) Solution: 1 det(A I) = so 7 and 1 = 2 4 2 6 = (1 )( 6 ) 8= 2 +5 14 = ( +7)( 2), = 2. (b) Compute an eigenvector ~v1 that spans the 1 -eigenspace of A and an eigenvector ~v2 that spans the 2 -eigenspace of A. (2 points) 8 2 4 1 1 Solution: A + 7I = ! , so ~v1 = . 4 1 0 0 4 1 2 1 2 2 A 2I = ! , so ~v2 = . 4 8 0 0 1 (c) Use parts (a) and (b) to diagonalize A. (Hint: Write A = P DP invertible and D is diagonal.) (1 point) A= 1 2 4 1 7 0 0 2 1 9 1 4 2 5 47 2 The eigenvalue is 10. 2 8 7 , where P is 2 1 2 3 2 1 5 4 5 4 (2) Show that the vector 2 is an eigenvector of the matrix 7 3 2 eigenvalue? (2 points) Solution: 1 2 8 7 3 3 15. What is its 6 32 3 2 3 2 3 2 3 3 1 5 4+9 10 1 5 4 5 4 5 4 5 4 1 2 = 7 + 16 3 = 20 = 10 25 6 3 2 + 14 + 18 30 3 Math 2270, Fall 2015 Instructor: Thomas Goller d dt (3) Bonus problem: Consider the derivative P3 (1 bonus point) (a) Use the standard basis for P3 to view compute a matrix A for S. Solution: Since dtd (1) = 0, coordinates gives us d (t) dt d dt = 1, 2 0 60 A=6 40 0 / 10 November 2015 P3 , which is a linear transformation. as a linear transformation R4 d 2 (t ) dt = 2t, and 1 0 0 0 3 0 07 7. 35 0 0 2 0 0 d 3 (t ) dt S / R4 , and = 3t2 , taking E- (b) Compute the eigenvalues and eigenvectors of A. Is A diagonalizable? Solution: 1 det(A so = 0 and n1 = 4. I) = 0 0 0 0 0 0 2 0 0 3 = 4 , 0 3 1 0 0 0 2 07 7 A 0 0 35 0 0 0 2 3 1 607 7 yields only one independent eigenvector ~v1 = 6 405, so A is not diagonalizable. 0 1 2 0 60 0I = 6 40 0 (c) What does your result in (b) tell you about how d dt acts on polynomials? Solution: The eigenvector ~v1 corresponds to the polynomial 1, so we see that the scalar polynomials are mapped to 0 by dtd . Since there are no other eigenvectors, we see that dtd does not map any other polynomial to a multiple of itself, which makes sense because dtd decreases the degree of the polynomials on which it acts.