(a) only (ii) and (iv) (b) only (ii) and (iii) (c) only (i) and (ii) (d) only (iv) (e) only (i) and (iii)

Transcription

1 . Which of the following are Vector Spaces? (i) V = { polynomials of the form q(t) = t 3 + at 2 + bt + c : a b c are real numbers} (ii) V = {at { 2 + b : a b are real numbers} } a (iii) V = : a 0 b is any real number b (iv) V = a b : a + b + c = 0 c (a) only (ii) and (iv) (b) only (ii) and (iii) (c) only (i) and (ii) (d) only (iv) (e) only (i) and (iii) (i) V is not a vector space since V does not contain the zero polynomial. (ii) We show that V is a subspace of P 2. Note that the zero polynomial is in V. For p(t) = at 2 + b q(t) = ct 2 + d and k R we check that p(t) + q(t) and kp(t) are still in V : p(t) + q(t) = (at 2 + b) + (a t 2 + b ) = (a + a )t 2 + (b + b ) kp(t) = (ka)t 2 + (kb). It follows that V is a subspace of P 2 and therefore a vector space. (iii) V is not a vector space. The vector v = is in V but v = is not in V. 0 0 (iv) V is a vector space. In fact V is the null space of the matrix A =. 2. Let A be the matrix A = t Find all possible values of t for which the system Ax = 0 admits only the trivial solution. (a) t 4. (b) t =. (c) t = t = and t = 2. (d) t can be any real number. (e) No such value of t exists. Since A is a 3 3 matrix it suffices to find t so that t is invertible or equivalently all values of t so that det(a) 0. In this case the determinant of A is 2 det(a) = det = t 4 2 t it follows that t 4. Note that we have computed the determinant of A by cofactor expanding along the second column.

2 3. The reduced echelon form of a 3 4 matrix A is as follows Which of the following statements is FALSE? (a) Nul A is a subspace of R 3. (b) The dimension of the column space of A is 2. (c) The row space of A is spanned by the vectors 0 2 T and 0 3 T. (d) The range of the linear transformation x Ax has dimension 2. (e) The dimension of the row space of A is the same as the rank of A. (a) Nul A is a subspace of R 4 not R 3. (b) There are two pivot positions in the reduced echelon form of A so the dimension of the column space of A which equals the rank of A is 2. (c) Since row operations preserve the row space the row space of A is spanned by the pivot rows. (In accord with the conventions of the book we write them as colum vectors by taking their transpose.) (d) The dimension of the range which equals the rank is 2. (e) This is true for all matrices. 4. Which of the following matrice(s) have characteristic polynomial ( λ) 2? 2 (i) 0 (ii) (iii) 0 (iv) 2 (a) only (i) (ii) and (iv) (b) only (ii) (c) only (i) and (ii) (d) only (iii) and (iv) (e) All the matrices have characteristic polynomial ( λ) 2. Solution.The correct answer is (a). λ 2 (i) det = ( λ) λ 2. λ 0 (ii) det = ( λ) λ 2. λ (iii) det = λ λ 2 +. λ (iv) det = λ(2 λ) + = 2λ + λ 2 λ 2 + = ( λ) 2.

3 5. λ = 2 is an eigenvalue for the matrix A = Which one of the following set forms a basis for the eigenspace of A corresponding to λ = 2? (a) 0 (b) 0 0 (c) 0 (d) (e) To find the eigenspace corresponding to λ = 2 we solve the homogeneous system (A 2I)x = 0. A 2I = The eigenspace Nul(A 2I) is spanned by Let R be the rectangle in R 2 with the vertices and let A = v Av. 3. Find the area of the image of R under the linear transformation (a) 8 (b) 8 (c) 0 (d) 2 (e) 4 Solution. The correct answer is (a). The area of the rectangle R is (3 ) ( 0) = 2 and det A = 4. The area of the image of R under v Av is the product of det A and the area of R which is 8.

4 7. Which of the following sets of vectors is a basis for the given vector space V? (i) { 2} V = R 2 (ii) V = R3 (iii) { t 2 t } V = P 2 (iv) {sin x cos x} V = C 0 (R) Note: In the questions above P 2 is the vector space of polynomials of degree less than or equal to 2 and C 0 (R) is the vector space of continuous functions on R. (a) Only (ii) (b) Only (iii) (c) Only (ii) and (iv) (d) Only (i) (iii) and (iv) (e) None of the sets is a basis. Solution.The answer is (a). (i) dim R 2 = 2. There are too many vectors. The given set is not a basis (ii) det = 2 3 = so the given set is linearly independent and therefore a basis. (iii) The dimension of P 2 is 3. There are too few elements. The given set is not a basis. (iv) C 0 (R) is infinite-dimensional. 8. For t > 2 consider the linear system: tx + x 2 = 3t x + (t )x 2 = Use Cramer s rule to find x. (a) x = 3t2 3t t 2 t (b) x = 2t t 2 t Solution.The answer is (a). By Cramer s rule x = (c) x = 3t2 t 2 t (d) x = t2 3t 3t 2 t 3t det t = 3t2 3t t t det 2 t. t (e) x = t2 t 3t 2 3t

5 9. Let A be a n n matrix that is NOT invertible. Which of the following statements are TRUE? (i) 0 is not an eigenvalue of A. (ii) There is a n n matrix B such that det(a T B) 0. (iii) Row A = R n. (iv) dim(nul A T ) = 0. (v) rank A < n. (a) Only (v). (b) Only (ii) and (v). (c) Only (i) and (v). (d) Only (i) (ii) (iii) and (iv). (e) All the statements are false. Solution.The answer is (a). (i) A is not invertible means that 0 is an eigenvalue of A. (ii) A is not invertible implies that A T is not invertible so det A T = 0 and det(a T B) = det A T det B = 0. (iii) A is not invertible so dim(row A) = rank(a) < n and Row A R n. (iv) A T is not invertible so dim(nul A T ) > 0. (v) A is not invertible so rank A < n. 0. Consider two bases for R 2. B = { 3 } 4 C = 5 Find the change-of-coordinates matrix P C B from B to C. Solution. Thus we have P = C B { 2 }. R R +R R 2 R 2 2R Given the matrix A = (a) Find all eigenvalues of A. (b) For each of the eigenvalue in part (a) find a basis for the eigenspace corresponding to that eigenvualue.

6 Solution. (a) The characteristic polynomial of A is p(λ) = det(a λi 3 ) = det λ 0 0 λ λ λ = ( λ) det = ( λ) 0 λ 2 ( + λ) The eigenvalues of A are λ = and λ 2 =. (b) For λ = we have A I 3 = and a basis of Nul(A I 3 ) is 0 0. For λ 2 = we have A + I 3 = and a basis for Nul(A + I 3 ) is Consider the matrix A = (i) Find the reduced echelon form of A. (ii) Find a basis for each of Row A Col A and Nul A.

7 Solution. (i) (ii) The set of pivot rows { } of the last matrix above is a basis of RowA. Since the book writes row vectors as (their transpose) column vectors the solution is that is a basis for Row A. However since the distinction between row and column vectors is just one of notation the first answer (given in the form of row vectors) was also accepted. The set of pivot colums of A is a basis for Col A. Writing the solutions of Ax = 0 in parametric form we have 3 2 x = x x Hence is a basis for Nul A