quiz3_1806_sol_s05 - Exam 3, Friday May 4th, 2005 Solutions...

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Unformatted text preview: Exam 3, Friday May 4th, 2005 Solutions Question 1. (a) The characteristic polynomial of the matrix A is ⎝ �2 1 32 9 1 3 = −(� − 1) � − −� + � − � + 2 16 16 4 and thus the eigenvalues of A are 1 with multiplicity one and 1 with multiplicity two. Clearly 4 the eigenvectors with eigenvalue 1 are the non-zero multiples of the vector (1, 1, 1) T . The remaining eigenvectors are all non-zero vector of the orthogonal complement of (1, 1, 1) T : they are the vectors ⎞ � a ⎠ b � , with (a, b) �= (0, 0) −a − b and an orthogonal basis for this vector space is ⎞ �⎞ � 1 1 ⎠−1� , ⎠ 1 � 0 −2 (b) For the matrix S we may choose the orthogonal matrix � ⎞1 1 1 ⎟ S=⎟ ⎠ � 3 1 � 3 1 � 3 � 2 −1 � 2 0 � 6 1 � 6 −2 � 6 � � � 1 and the matrix � is then the diagonal matrix with entries 1, 4 , and 1 along the diagonal. 4 We have ⎞ � 1 0 0 ⎛ ⎜ ⎟ � � lim Ak = S lim �k S −1 = S ⎟ 0 0 0 � S T ⎠ k �� k �� 000 and therefore ⎟ lim Ak = ⎟ ⎠ k �� ⎞ 1 � 3 1 � 3 1 � 3 00 � � 1⎟ � 0 0 � S T = ⎟ 1 1 1 � � 3⎠ 00 1 1 1 � ⎞ 1 1 1 � 1 (c) Any r < 4 is such that A − r I is positive definite. Since we want r to be positive, we may choose r = 1 . 8 1 Any 1 < s < 1 is such that A − sI is indefinite. We may choose s = 2 . 4 Any 1 < t is such that A − tI is negative definite. We may choose t = 2. 1 1 (d) The singular values of B are 1, 2 and 2 . 1 Question 2. The trace of A equals the sum of the eigenvalues, which is zero. We deduce that the entry in the second row and second column is −a. Similarly, the determinant of A equals the product of the eigenvalues, which is -1. We deduce that the entry in the second row and first column is 1 − a2 . Thus we have ⎝ � a 1 A= 1 − a2 −a (b) The matrix A has two independent eigenvectors since it has two distinct eigenvalues. (c) The only choices of a giving orthogonal eigenvectors are the ones for which A is symmetric. This implies a = 0. If a �= 0, then A does not have orthogonal eigenvectors. (d) For any choice of a the matrix A has exactly one eigenvalue 1 and exactly one eigenvalue -1. Thus the Jordan canonical form of A is always ⎝ � 10 0 −1 independently of what a is. Question 3. (a) The general solution to the differential equation u(t) = c1 e�1 t x1 + c2 e�2 t x2 + c3 e�3 t x3 where c1 , c2 , c3 are arbitrary constants. (b) Since the vectors x1 , x2 , x3 are independent, they form a basis for R3 . It follows that we may write any vector u0 � R3 as a linear combination of these vectors: u0 = a1 x1 +a2 x2 +a3 x3 . Repeatedly applying the matrix A we obtain u k = A k u 0 = � k a 1 x1 + � k a 2 x2 + � k a 3 x3 1 2 3 If we want the limit as k goes to infinity of the vectors uk to be zero, then all the limits lim �k must b e zero. It follows that we necessarily have −1 < �i < 1, for all i’s. i du dt = Au is 2 ...
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This note was uploaded on 01/14/2011 for the course EECS 18.06 taught by Professor Strang during the Spring '05 term at University of Michigan.

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