practiceFinalb

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern