Midterm 2 Solutions, MATH 54, Linear Algebra and Differential Equations, Fall 2014
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1 Name (Last, First): Student ID: Circle your section: 2 Shin 8am 7 Evans 22 Lim pm 35 Etcheverry 22 Cho 8am 75 Evans 23 Tanzer 2pm 35 Evans 23 Shin 9am 5 Latimer 24 Moody 2pm 8 Evans 24 Cho 9am 254 Sutardja Dai 25 Tanzer 3pm 26 Wheeler 25 Zhou am 254 Sutardja Dai 26 Moody 3pm 6 Evans 26 Theerakarn am 79 Stanley 27 Lim 8am 3 Hearst 27 Theerakarn am 79 Stanley 28 Moody 5pm 7 Evans 28 Zhou am 254 Sutardja Dai 29 Lee 5pm 3 Etcheverry 29 Wong 2pm 3 Evans 22 Williams 2pm 289 Cory 2 Tabrizian 2pm 9 Evans 22 Williams 3pm 4 Barrows 2 Wong pm 254 Sutardja Dai 222 Williams 2pm 22 Wheeler If none of the above, please explain: This is a closed book exam, no notes allowed. It consists of 6 problems, each worth points. We will grade all 6 problems, and count your top 5 scores. Problem Maximum Score Your Score Total Possible 5
2 Problem ) Decide if the following statements are ALWAYS TRUE or SOMETIMES FALSE. You do not need to justify your answers. Write the full word TRUE or FALSE in the answer boxes of the chart. (Correct answers receive 2 points, incorrect answers or blank answers receive points.) Statement Answer TRUE FALSE FALSE FALSE TRUE ) The matrices A = 2) The matrices A = [ ] and B = [ ] and B = [ ] are similar. [ ] are similar. 3) For 2 2 matrices A and B, if v is an eigenvector of AB, then Bv is an eigenvector of A. 4) If a 3 3 matrix A is diagonalizable with eigenvalues ±, then it is an orthogonal matrix. 5) If u v = u + v, then u and v are orthogonal. 2
3 Problem 2) Indicate with an X in the chart all of the answers that satisfy the questions below. You do not need to justify your answers. It is possible that any number of the answers satisfy the questions. (A completely correct row of the chart receives 2 points, a partially correct row receives point, but any incorrect X in a row leads to points.) (a) (b) (c) (d) (e) Question X X X Question 2 X X Question 3 X X X Question 4 X X Question 5 X X X X ) Which of the following conditions guarantees an n n real matrix A is diagonalizable with real eigenvalues? a) Every eigenvalue of A has an eigenvector. b) There is a basis of R n consisting of real eigenvectors for A. c) det(a λi n ) = λ n λ n and dim N ul(a) =. d) det(a λi n ) = λ n λ n 2 and dim N ul(a) = n 2. e) The inverse of A is diagonalizable with real eigenvalues. 2) For what h is the matrix h 2 2h 2h h h 2 diagonalizable with real eigenvalues? a) h = 2 b) h = c) h = d) h = e) h = 2 3
4 3) Which of the following linear transformations T : P 2 P 2 have rank? a) T (p(x)) = p (x) b) T (p(x)) = p (x) c) T (p(x)) = ( + x)p (x) d) T (p(x)) = ( + x)p (x) e) T (p(x)) = ( + x)p() 2 2 4) Which of the following are a basis B of R 3 so that for x = 3 we have [x] B =? a), 3, 3 b),, 3 c) 2,, d) 3, 9, e),, 2 5 5) Which of the following linear transformations R 2 R 2, x Ax are given by an orthogonal matrix A? a) Reflection across the line x = y. b) Rotation by π/4 about the origin. c) A shear transformation fixing the line y =. d) Reflection across the line x = y followed by reflection across the line x =. e) Scaling by 2 followed by rotation by π/4 about the origin followed by scaling by /2. 4
5 Problem 3) a) (4 points) Find the eigenvalues and a basis consisting of eigenvectors of A = Solution: det(a λi) = λ( λ)( 2 λ) so eigenvalues are,, 2. The corresponding eigenvectors are v =, v 2 =, v 3 = b) (3 points) Find the coordinates of the vector v = with respect to the basis of eigenvectors. Solution: so coordinates are v = + ( ) + 5
6 c) (3 points) Calculate A 24 v. Solution: A 24 v = A 24 (v v 2 + v 3 ) = v 2 + ( 2) 24 v 3 = + ( 2) 24 + ( 2) 24 6
7 Problem 4) a) (4 points) Calculate the matrix [T ] of the linear transformation p() T : P 2 R 3 T (p(x)) = p () p () p () + p () with respect to the basis B = {, + x, + x + x 2 } of P 2 and the standard basis of R 3. Solution: Columns of [T ] result from applying T to the basis vectors of B and expanding in terms of the standard coordinate basis. 2 3 [T ] = b) (4 points) Find bases of P 2 and R 3 such that the matrix of T satisfies [T ] = Solution: Take for example the standard basis, x, x 2 of P 2 and the resulting basis of images T () =, T (x) =, T (x 2 ) =
8 c) (2 points) Find bases of P 2 and R 3 such that the matrix of T satisfies [T ] = Solution: Take the same bases as in part b). 8
9 Problem 5) Consider the subspace W of R 4 spanned by u = 2 v = 2 4 a) (4 points) Find a nonzero vector w in W orthogonal to u. Solution: Set w = au + bv. We want w u =. We find w u = 9a + 9b. So for example take a =, b = and so w = u v = 2 2 b) (3 points) Find the orthogonal projection of the vector 3 y = 2 to the subspace W. Solution: We calculate /9 ŷ = y u u u u + y w w w w = 9 u w = 7/9 2/9 6/9 9
10 c) (3 points) Find the orthogonal projection of the vector y to the orthogonal subspace W. Solution: Take 3 /9 26/9 z = y ŷ = ( 7/9) 2 2/9 = 2/9 6/9 6/9 7/9
11 Problem 6) ( points) Fill in the blanks (each worth /2 a point) in the proof of the following assertion. Assertion. If a 2 2 matrix A satisfies det(a λi) = λ 2, then A 2 =. Proof. Since det(a λi) = λ 2, the only eigenvalue of A is. There must be a corresponding eigenvector, which we will call v, because det(a) = implies A is not invertible, and therefore N ul(a) must be nontrivial. Choose any w linearly independent from v. Thus the pair v, w is a basis, which we will call B, because v, w must also span R 2. Thus there exist a, b so that Aw = av + bw. The matrix of A with respect to B is then [A] B = a b We see that b =, since the diagonal entries of any triangular matrix are its eigenvalues. Finally, let P B be the matrix with columns v, w. Then A = P B [A] B P B. Since it is easy to see that [A] 2 B =, we also find A 2 = (P B [A] B P B )2 = P B [A] 2 BP B =.
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