1. Let A = (a) 2 (b) 3 (c) 0 (d) 4 (e) 1

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1 . Let A =. The rank of A is (a) (b) (c) (d) (e). Let P = {a +a t+a t } where {a,a,a } range over all real numbers, and let T : P P be a linear transformation dedifined by T (a + a t + a t )=a +9a t If T p(t) = λp(t) for some non-zero polynomial and some real number λ, then p(t) is called an eigenvector corresponding to λ. The linear transformation T has an eigenvector: (a) (b) (c) t (d) t +9t (e) t. Let {λ,λ } be two eigenvalues of equal to 6. Then the product of two eigenvalues, λ λ, is (a) 8 (b) 8 (c) (d) (e) 7. Let b =,Letb =, Let x = and B = {b, b } Find [x] B. (a) (b) (c) (d) (e). The matrix A = has a complex eigenvector: (a) (b) (c) i i 6. Let A = 6. The eigenvalues of A are (d) (e) i

2 (a),, (b), ± (c),, (d),, (e),, 7. Let S be the parallelogram determined by the vectors b = and b = 99 A =. Compute the area of the images of S under the mapping v Av. and let (a) (b) 99 (c) (d) (e) 8. Let A =. The complex eigenvalues of A are (a) ± i (b) ± i (c) ±i (d) (e) ±i 9. For what value(s) of h will y be in the subspace spanned by {v, v, v }, if v = v =, v and y = 6. h, (a) (b) (c) (d) 6 (e) 9. Let b =, b =, c =, c =, B = {b, b } and C = {c, c }. If P = [b ] C, [b ] C is the change-of-coordinates matrix from B to C, then find P. (a) (b) 9 6 (c) (d) (e) 9 7 9a 8b. Find a matrix A such that W = Col(A) where W = a +b a all real numbers. and {a, b} range over

3 (a) 9 8 (b) (c) 9 8 (d) 9 8 (e). Let S =,. Then the subset S (a) is orthogonal to (b) a basis of R (c) spans R (d) a linearly dependent subset (e) a linearly independent subset. Let A = 8 9 Use Cramer s rule to solve Ax =. 7. Let P = {a +a t+a t } where {a,a,a } range over all real numbers, and let T : P P be a linear transformation given by T (a +a t+a t )=a +a t. Suppose that B = {,t,t } is a basis of P and C = {,t} is a basis of P. () Find a matrix A such that [T v] C = A[x] B. () Find Nul(A) and Col(A).. Let A = 7. Find lim A k. (Hint: Find the Diagonalization D of A use the formula k A k = PD k P.

4 Solutions. Reduce to echelon form: two pivots so rank is There are. With the basis B = {,t,t }, the matrix for T is 9. Because the matrix is triangular, the characteristic polynomial is λ and hence the eigenvalue is and the eigenvector is, so = + t +t is an eigenvector with eigenvalue.. The product of the roots, with multiplicities, of any polynomial with leading coefficient is ( ) n times the product of the roots where n is the degree: (t λ )(t λ ) (t λ n )= t n + +( ) n (λ λ n ). In this case λ λ = det A so the answer is 8.. We need to solve Ay = x where A =. So reduce to reduced row echelon form Hence the solution is [x] 9 B =. The characteristic equation is det λ λ =( λ) +9 = 6 8λ+λ +9 = λ 8λ+. The roots are λ = 8 ± 6 = ± i. The eigenspace for + i is the null space for i and an eigenvector is. i i

5 6. λ The characteristic equation is det 6 λ =( λ) det 6 λ λ λ det λ + det 6 λ =( λ) ( 6 λ)( λ)+9 ( ( λ)+9 + ( 6 λ) and good luck to you! OR The sums of the five answers are all the different and the sum of the eigenvalues is the trace of the matrix + ( 6) + =. Hence the only choice for the eigenvalues in the given answers is,,. 7. The area is the area of S times the absolute value of the determinant of A, which is. The area of S is det [b b ] = det = 9=. 8. The characteristic equation is λ traλ+det A = λ 6λ+ with roots λ = 6+ 6 ± i. = 9. We are looking for the h such that the equation Ax = y with A = so we need to reduce the following matrix to reduced row echelon form: = 6 8. h h 8 h h To have a solution, h = We need to write b = = ac + bc = a + b or solve x = so the first column of P is. 7 so the second column of P is. 7

6 6 Hence P =.. 9a 8b a +b = a 9 + b 8 so a matrix is 9 8 a. Since there are only two vectors they can not be a basis for R. They can not even span R. If the two are dependent, one is a multiple of the other and this is clearly not the case. Hence they are independent. The dot product is so they are not orthogonal.. To use Cramer s rule we need to compute four determinants: det = det 8 7 ( 9) det 7 += + 7 = det 8 9 = det 7 7 += det 9 = + det = and det 8 = det 7 7 += Hence the solution vector is x =.. A is a matrix and the i th column is found by working out T on the i th basis element of B.

7 Column : T () = so. Column : T (t) =so. Column : T (t )=so. Hence A =. The matrix A is already in echelon form and there are two pivot positions. Hence Col(A) has dimension and so is all of P. The space Nul(A) has dimension, non-zero vector in the null space so it is a basis for it. is clearly a. The characteristic equation for A is λ trace(a)λ + det(a) =λ 8λ + =(λ )(λ ). Hence the eigenvalues are and. The eigenspace for is the null space for and this is spanned by. The eigenspace for is the null space for and this is spanned by If P = and D =, then A = PDP. Note lim D k = so k = = =. lim k Ak =

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