Exam 2 Solution

# 1 0 2 for v for 2 1 0 so 2 v is 1 0 2 3 0 6 has

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Unformatted text preview: 0, 1 1 0. 1 By the Diagonalization Theorem and Theorem 7 of Ch 5 (dimension of each eigenspace equals the multiplicity of the corresponding eigenvalue), is diagonalizable as with 1 0 0 0 2 0 0 0 and 2 13 01 20 1 0 1 2. [10 pts] Consider a dynamical system described by the difference equation x x 0, 3/4 1/4 1 3 where . The vectors v and v are eigenvectors of corresponding 1/4 17/12 3 1 2 . to eigenvalues 2/3 and 3/2, respectively. Let x be a solution of the difference equation for x 2 Find a formula for x that does not directly involve the matrix , and describe in words what happens to ∞. x as , s.t. x Because 1, Find v1 v. 1, goes to ∞ for large . v 312 ~ 132 Row reduce [v1 v x 0.4 v1 goes to 0 for large k, and because So for large k, x So x v for solution x v1 0.8 132 1 ~ 312 0 v , and for large k, x is a multiple of v , and the growth rate is 3/2. 3 10 0.8 2 132 1 ~ ~ 8 0 1 .8 0 0 1 .4 .8 1 . That is, as k goes to infinity, x 3 3. [8 pts] Consider p 1 3 ,p 2 3 ,p 1 2. (i) Use coordinate vectors to show that these polynomials form a basis B for (ii)...
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## This note was uploaded on 01/17/2014 for the course ESE 309 taught by Professor Staff during the Fall '08 term at Washington University in St. Louis.

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