practiceFinalb - Math 310 Final Exam(Solutions Prof S Smith...

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Unformatted text preview: Math 310: Final Exam (Solutions) Prof. S. Smith: Wed 8 Dec 1993 Problem 1: (a) Find the eigenvalues of A = OMH Ol—‘N GOOD (b) Is A diagonalizable? (Give reason, not necessary to do diagonalization). (C) Is A positive definite? Give two methods of deciding. (a) We see det(A—$I) = (3—3:)[(1—a:)2—2-2] = (3—3:)(302—233—3) = (3—$)(:1:—3)($+1). So eigenvalues —1, 3, 3. (b) Yes. Get linearly independent eigenvectors (1,1,0) and (0,0,1) for 3. So With eigenvector (1, —1,0) for —1, have basis of eigenvectors. (c) No. From above, not eigenvalues are positive. Alternatively, the three “upper lef ” determinants are 1, —3, —9, not all positive. Problem 2: Let a linear transformation T : R3 —> R3 be defined by T(v1, v2, v3) = (3111 + 2112 + 113, 2111 + 112,122). Give the matrix (in the standard basis) for T. Compute T(1, 0, 0) = (3, 2, 0) and T(0,1,0) : (2,1,1) and T(0,0, 1) = (1,0,0). 3 2 1 Use as columns to get 2 1 0 . 0 1 0 OJIl-‘OJIN 1 Problem 3: Let M be the Markov matrix ( g 3 (a) Find the “steady state” eigenvector for M. (Components of vector should add to 1). (b) Diagonalize M and use this to get an expression for the power M n. (a) For eigenvalue 1, M — 1] 2 % < _3 _§ ), so eigenvector is span of (1, 1)T. Hence (%, is steady-state vector. (b) Compute det(M — x1) = (a: — %)2 — g2 = 3:2 — gm — g = (x — 1)($+ 80 other eigenvalue is —%; from M + %I = g < 3 g ) get eigenvector (1,—1)T. 1—1 71—1)“ <—%>g>%(i —i> Hi _1>=%<1t2:;:1;;::s:>- UsingS = (1 1 >, get 3—1MS = A so M 2 SAS‘1 hence M“ = S(A”)S‘1. ) Thus M” = < 1 i g) =(1_E: Problem 4: Let A : HHH 1—11—11_\ 1 1 . I give you that the eigenvalues are 3,0,0. 1 (a) Find eigenvectors for these eigenvalues. (b) Note A is symmetric. So find orthogonal S with S’lAS diagonal. (Remember this means the columns of S must be orthonormal). (c) Give the matrix P for projection on your 3-eigenvector. 1 0 —1 (a) A — 31 row—reduces to 0 1 —1 , so a 3—eigenvector is (1, 1,1)T. 0 0 0 1 1 1 A — OI : A row-reduces to 0 0 0 , U 0 0 so independent O-eigenvectors are (1, —1,0)T and (1,0, —1)T. (b) We need to apply Gram-Schmidt to each eigenspace. For 3, just divide by length to get i(1,1,1)T. x/§ For 0, first is (1, —1,0)T and second is (column form of) (1,0, —1) — (1, ammu, —1,0)(1,0, —1)T = (1,0, —1) — §(1, —1,0) = 1(1, 1, —2). 1 2 O c l D1V1de by lengths to get E( , «a —1,0)T and i(1,1,—2)T. L L L SouseSz ? —E E . E 0 ‘W 1 1 1 (C) i 1 i ...
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