Abstract Linear Algebra, Fall Solutions to Problems III


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1 Abstract Linear Algebra, Fall Solutions to Problems III 1. Let P 3 denote the real vector space of all real polynomials p(t) of degree at most 3. Consider the linear map T : P 3 P 3 given by T p(t) = tp (t). Determine the matrix of T with respect to the basis 1, t, t 2, t 3 of P 3? What is the matrix of T 5 with respect to this basis? We have T 1 = =. +.t +.t 2 +.t 3, T t = t =. + 1.t +.t 2 +.t 3, T t 2 = 2t 2 =. +.t + 2.t 2 +.t 3, T t 3 = 3t 3 =. +.t +.t t 3, and thus the matrix M(T ) of T with respect to the basis 1, t, t 2, t 3 is M(T ) = From class, M(T 5 ), the matrix of T 5 with respect to this basis, is M(T ) 5 = Let F be a field. For 1 i, j 2, write ij for the 2 2 matrix (over F ) with 1 in the ij position and in all other positions. a. Show that 11, 12, 21, 22 is a basis of M 2 (F ). b. Determine the matrix with respect to this basis of the linear map A A : M β 2 (F ) M 2 (F ). c. Determine the matrix with respect to this basis of the linear map A A : M β 2 (F ) M 2 (F ). a. We have [α ] β = α γ δ 11 + β 12 + γ 21 + δ 22, for any α, β, γ, δ F. This shows that 11, 12, 21, 22 generate M 2 (F ). It also shows that if [ ] α 11 + β 12 + γ 21 + δ 22 =,
2 2 then α = β = γ = δ =, i.e., 11, 12, 21, 22 are linearly independent. Thus 11, 12, 21, 22 is a basis of M 2 (F ). For simplicity, we write B from now on for this (ordered) basis. b. We have Similarly, [ ] [ ] [ ] α 1 α 11 = β β = = α = β β 12, [ ] α 21 = α β 21, 22 = β β 22. It follows that the matrix with respect to B of the given linear map is α β α. β c. As above, β 11 = α 11, β 12 = α 12, β 21 = β 21, β 22 = β 22. Hence the matrix with respect to B of the given linear map is α α β. β 3. Let A be an m n matrix over a field F. We view the elements of F n and F m as column vectors of size n and m (respectively). Consider the linear map L A : F n F m given by L A (X) = AX, X F n.
3 3 Recall that the standard basis of F n consists of the vectors =., 2 =.,..., n =.. 1 Of course, we also have the corresponding standard basis of F m. a. What is the matrix of L A with respect to these bases? b. Let B be an n p matrix over F. As above, B defines a linear map L B : F p F n. What is the matrix of the composition L A L B : F p F m with respect to the standard bases of F p and F m? c. Now let C be a p q matrix over F. We again have a linear map L C : F q F p. xplain why (L A L B )L C = L A (L B L C ). Deduce that matrix multiplication is associative, i.e., (AB)C = A(BC). a. Throughout we write M( ) for the matrix of any linear map : F F with respect always to the standard bases of these spaces. By direct calculation, the ith column of A. Thus a 1i a 2i L A ( i ) = A i =., a mi L A ( i ) = a 1i 1 + a 2i a mi m, where, for simplicity, we use the same symbols for the standard basis vectors in F n and F m. It follows that the ith column of the matrix M(L A ) is simply the ith column of A. Hence M(L A ) = A. b. We have M(L A L B ) = M(L A )M(L B ) (from class) = AB (by part a). c. Recall that composition of functions is associative. Indeed, suppose we have A h B g C f D. That is, we re given functions f : C D, g : B C, h : A B between sets A, B, C, D. Then (fg)h : A D satisfies ((fg)h)(a) = (fg)(h(a)) = f(g(h(a))), for all a A.
4 4 Similarly, the function f(gh) : A D satisfies (f(gh))(a) = f((gh)(a)) = f(g(h(a))), for all a A. Thus (fg)h and f(gh) coincide at each a A and so (fg)h = f(gh), i.e., composition of functions is associative. In particular, composition of linear maps is associative and hence (L A L B )L C = L A (L B L C ). Therefore M((L A L B )L C ) = M(L A (L B L C )) ( ). We have M((L A L B )L C ) = M(L A L B )M(L C ) (from class) = (AB)C (by parts a and b). In the same way, M(L A (L B L C )) = A(BC). Hence ( ) implies (AB)C = A(BC), as required. Here s an alternative more direct solution. For any column vector X in F q, In the same way, for any X F q, (L A L B )L C X = (AB)C X. L A (L B L C ) X = A(BC) X, and thus (AB)C X = A(BC) X, for all X F q. Take X = i, the ith standard basis vector in F q, so that (AB)C i = A(BC) i. As noted in the solution to part a, this says that the ith columns of (AB)C and A(BC) are equal, whence once more (AB)C = A(BC). 4. Let V be a vector space and let s : V V and t : V V be linear maps such that st = 1 V, the identity map on V. a. Show that s is surjective and t is injective. b. Suppose V has finite dimension. Show that ts = 1 V. a. For any v V, s(t(v)) = v. In other words, s(w) = v for w = t(v), and thus s is surjective. Suppose t(v 1 ) = t(v 2 ), for v 1, v 2 V. Then st(v 1 ) = st(v 2 ), i.e., v 1 = v 2 (since st = 1 V ), and thus t is injective. b. Since V is finitedimensional and s is surjective, we know from class that s is bijective. Hence s 1 exists and so st = 1 V implies s 1 st = s 1 1 V. Thus t = s 1 and as required. ts = s 1 s = 1 V,
5 5 5. Let P denote the real vector space of all real polynomials. Consider the linear maps D : P P and I : P P given by Dp(t) = p (t), Ip(t) = p(s) ds. Show that DI = 1 P, the identity map on P, but that ID 1 P. This shows that the conclusion of 3b fails for infinite dimensional spaces. For p(t) in P, DIp(t) = d dt p(s) ds = p(t) (by the fundamental theorem of calculus), and so DI = 1 P. Let c be a nonzero constant. Then IDc =, so IDc c, and thus ID 1 P. In general, for p(t) in P, IDp(t) = and so IDp(t) = p(t) if and only if p() =. p (t) dt = p(t) p(), 6. Let t : V V be a linear map and suppose v 1,..., v n are vectors in V such that t(v 1 ),..., t(v n ) generate V. Does it follow that t is invertible? Let v V. By hypothesis, there exist scalars α 1,..., α n such that v = α 1 t(v 1 ) + + α n t(v n ) = t(α 1 v α n v n ). Hence v is in the image of t and so t is surjective. Observe that V is finitedimensional as it is generated by the finite list of vectors t(v 1 ),..., t(v n ). The surjective linear map t is therefore bijective and thus has an inverse. In other words, t is invertible.
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