Solutions to Homework 1.
Math 110, Fall 2006. Prob 1.2.1. (a) True this is axiom (VS 3). (b) False we proved the zero vector is unique. (c) False: take x to be the zero vector, and take any scalars a and b. (d) False: take a to be the zero scalar and any
Solutions to Homework 11.
Math 110, Fall 2006. Prob 5.4.1. (a) False, cfw_0 and the whole space are T -invariant for any T . (b) True (Theorem 5.21). (c) False: just take w to be a nonzero multiple of v, then the corresponding T -cyclic subspaces are the
Solutions to Homework 10.
Math 110, Fall 2006. Prob 5.1.3. (a) The eigenvalues are 1 and 4, with the eigenspaces E1 = spancfw_[1 1]t , E4 = t spancfw_[2 3] . The vectors [1 1]t , [2 3]t form a basis for IR2 . The matrix Q diagonalizes A, where Q= 1 1 2 3
Solutions to Homework 9.
Math 110, Fall 2006. Prob 4.3.10. Since det(AB) = det(A) det(B) for any two square matrices of the same order, this implies, by induction, that (det M )k = det(M k ) for all k IN. Since the determinant of the zero matrix is zero,
Solutions to Homework 8.
Math 110, Fall 2006. Prob 4.1.1. (a) False, it is 2-linear. (b) True. (c) False, A is invertible if and only if det(A) = 0. (d) False, it is the absolute value of that determinant. (e) True (proved in this section). Prob 4.1.2. (a
Solutions to Homework 6.
Math 110, Fall 2006. Prob 2.7.1. (a) True. (b) True. (c) False: the roots of the auxiliary polynomial give frequencies of solutions. (d) False: linear combinations of these are solutions too. (e) True: directly from linearity and
Solutions to Homework 4.
Math 110, Fall 2006. Prob 2.3.13. Let A = [aij ]. Then At = [aji ], and
ajj = tr(At ).
The elements of A = [aij ], B = [bij ], AB = [cij ] and BA = [dij ] are connected by the formulas
Solutions to Homework 3.
Math 110, Fall 2006. Prob 2.1.10. By the linearity of T , we use the fact (1, 0) + 3(1, 1) = (2, 3) to obtain T (2, 3) = T (1, 0) + 3T (1, 1) = (1, 4) + 3(2, 5) = (5, 11). The map T is 1 1 as we see that T (a(1, 0) + b(1, 1) = a(1
Solutions to Homework 2.
Math 110, Fall 2006. Prob 1.4.4. (a) Yes, since the linear system a+b 2a + 3b -a a-b has a solution a = 3, b = -2. (b) No, the corresponding linear system has no solution. (c) Yes, the corresponding linear system has a solution a
Solutions to Homework 12.
Math 110, Fall 2006. Prob 6.1.1. (a) True, directly by definition. (b) True, this is also required by the definition. (c) False, it is linear in the first component and conjugate linear in the second component. (d) False, there a