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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—$I) = (3—3:)[(1—a:)2—22] = (3—3:)(302—233—3) = (3—$)(:1:—3)($+1). So eigenvalues —1, 3, 3.
(b) Yes. Get linearly independent eigenvectors (1,1,0) and (0,0,1) for 3.
So With eigenvector (1, —1,0) for —1, have basis of eigenvectors. (c) No. From above, not eigenvalues are positive.
Alternatively, the three “upper lef ” determinants are 1, —3, —9, not all positive. Problem 2: Let a linear transformation T : R3 —> R3 be deﬁned by T(v1, v2, v3) = (3111 + 2112 + 113, 2111 + 112,122).
Give the matrix (in the standard basis) for T. Compute T(1, 0, 0) = (3, 2, 0) and T(0,1,0) : (2,1,1) and T(0,0, 1) = (1,0,0). 3 2 1
Use as columns to get 2 1 0 .
0 1 0 OJIl‘OJIN 1
Problem 3: Let M be the Markov matrix ( g 3 (a) Find the “steady state” eigenvector for M. (Components of vector should add to 1).
(b) Diagonalize M and use this to get an expression for the power M n. (a) For eigenvalue 1, M — 1] 2 % < _3 _§ ),
so eigenvector is span of (1, 1)T. Hence (%, is steadystate vector.
(b) Compute det(M — x1) = (a: — %)2 — g2 = 3:2 — gm — g = (x — 1)($+ 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 3eigenvector. 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 rowreduces to 0 0 0 ,
U 0 0 so independent Oeigenvectors are (1, —1,0)T and (1,0, —1)T.
(b) We need to apply GramSchmidt 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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 Spring '08
 Smith
 Linear Algebra, Eigenvalues, linearly independent eigenvectors, Prof. S. Smith, Diagonalize M

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