Math 314H Solutions to Homework # 1

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1 Math 314H Solutions to Homework # 1 1. Let β = {1 + x, 1 + x, x + x } be a subset of P. (a) Prove that β is a basis for P. Let δ = {1, x, x } be the standard basis for P and consider the linear transformation T : P R 3 defined by T (f) = [f] δ, where [f] δ is the coordinate vector of f with respect to δ. Now, β is a basis for P if and only if T (β) = 1, 0, is a basis for R 3. (This follows from Theorem 8 on page 44. For further explanation, see Exercises 5 and 6 on page 49.) To check T (β) is a basis for R 3, it suffices to put the three columns in 3 3 matrix and show that the rref of this matrix is the identity matrix. (This computation is trivial, so I won t reproduce it here!) (b) Find the coordinate vector of 7 5x + 3x with respect to β. We need to find scalars c 1, c, c 3 such that c 1 (1 + x) + c (1 + x ) + c 3 (x + x ) = 7 5x + 3x. Comparing coefficients, this leads to the system c 1 + c = 7 c 1 + c 3 = 5 c + c 3 = 3. Solving this system, we get c 1 = 1, c = 15, c 3 = 9. Thus, the coordinate vector of 7 5x + 3x with respect to β is Let S = {1 x + 3x 3, 1 + x + x x 3, 3 + 5x + 8x + x 3, 4x + 5x + 3x 3 } be a subset of P 3. (a) Find a basis for Span(S). As in Problem 1, we use the coordinate mapping T : P 3 R 4 defined by T (f) = [f] δ, where δ = {1, x, x, x 3 }. It then suffices to find a basis for Span(T (S)), where T (S) is the set of coordinate vectors of the vectors in S: T (S) = , 1, 5 8,

2 To find a basis for the span of T (S), we write these vectors as the columns of a matrix and find its rref (which I did using Maple): From the rref, we see that our original matrix has pivots in the first, second, and fourth columns. Therefore, the first, second, and fourth polynomials form a basis for Span(S): {1 x + 3x 3, 1 + x + x x 3, 4x + 5x + 3x 3 }. (b) Decide whether f = x + 48x + 3x 3 is in Span(S) and if so, find the coordinate vector of f with respect to the basis you found in part (a). To see if f is in Span(S), we need to see if there exist scalars c 1, c, c 3 such that c 1 (1 x +3x 3 )+c ( 1+x+x x 3 )+c 3 (4x+5x +3x 3 ) = 9+35x+48x +3x 3. Comparing coefficients as we did in Problem 1, we get a system of equations whose augmented matrix is: which reduces to Since the system is consistent, f is in the span of S; its coordinate vector with 10 respect to the basis we found in part (a) is Let f : U V be a linear transformation of vector spaces. Prove that im f is a subspace of V. Since f(0) = 0, we see that 0 im f. Let v 1, v be vectors in im f. Then v 1 = f(u 1 ) and v = f(u ) for some u 1, u U. Then v 1 +v = f(u 1 )+f(u ) = f(u 1 +u ). This means that v 1 + v is in im f, since v 1 + v = f(w), where w = u 1 + u. Similarly, let c R be a scalar. Then cv 1 = cf(u 1 ) = f(cu 1 ). Thus, cv 1 im f. This proves im f is a subspace of V. 4. Let S and T be linear transformations from R to R which are defined by T ((a 0, a 1, a,... )) = (a 1, a, a 3,... ), S((a 0, a 1, a,... )) = (0, a 0, a 1, a,... ) and for all sequences (a 0, a 1, a,... ) in R. Furthermore, let I : R R be the identity map; i.e., I((a 0, a 1, a,... )) = (a 0, a 1, a... ).

3 (a) Find the image and kernel of T. Since T ((0, a 0, a 1, a,... )) = (a 0, a 1, a,... ), we see that every vector in R is in the image of T ; i.e., im T = R. Suppose (a 0, a 1,... ) ker T. Then T (a 0, a 1,... ) = (a 1, a,... ) = (0, 0, 0,... ). Therefore, a i = 0 for all i 1. Hence, ker T = {(a 0, 0, 0, 0, a 0 R}. (b) Find the image and kernel of S. Clearly, ker S = {(0, 0, 0,... )}, since if S((a 0, a 1, a,... )) = (0, 0, 0,... ) then a i = 0 for all i. Also, im T = {(0, a 1, a, a 3,... ) a i R}. (c) Is ST = I? (I.e., is S(T (a)) = a for all a R?) No. For example, (ST )((1, 0, 0,... )) = S(T ((1, 0, 0... ))) = S((0, 0, 0,... )) = (0, 0, 0,... ). (d) Is T S = I? Yes, since (T S)((a 0, a 1, a,... )) = T (S((a 0, a 1, a,... ))) = T ((0, a 0, a 1,... )) = (a 0, a 1, a,... ). (e) What does your answers to parts (c) and (d) tell you about the left invertibility and right invertibility of linear transformations of infinite dimensional vector spaces? It shows that a linear transformation can have left inverse which is not a right inverse. This stands in contrast to the finite dimensional case: if T and S are linear transformations from V to V where V is a finite dimensional vector space and T S = I, then ST = I also. (This follows from what we have proved about square matrices.) 5. Let S = { sin x, 3 cos x, cos x} be a subset of C(R), the vector space of continuous functions from R to R. Is S linearly independent or dependent? Justify your answer. How about {x, sin x, cos x}? Again, justify your answer. From the trig identity cos x = cos x sin x, we get 3( sin x) (3 cos x)+6(cos x) = 0. Hence, S is linearly dependent. The second set is linearly independent. To prove this, assume we have scalars c 1, c, c 3 such that c 1 x + c sin x + c 3 cos x = 0. This equation must hold for all x. Letting x = 0, we have c 3 = 0. Thus, the equation reduces to c 1 x + c sin x = 0. Letting x = π, we see that c 1 = 0. This leaves us with c sin x = 0. Letting x = π, we obtain c = Let B =. Let V = M 4 (R), the vector space of matrices with real entries. Define f : V V by f(a) = BA.

4 (a) Prove that f is a linear transformation. By properties of matrix multiplication, we have f(a + C) = B(A + C) = BA + BC = f(a) + f(c) for any A, C V. For any scalar c R and A V, we have f(ca) = B(cA) = c(ba) = cf(a). Thus, f is a linear transformation. (b) Find a bases for the kernel and image of f. a b We first find a basis for ker f. Let A =. Then c d A ker f f(a) = 0 Thus, A ker f if and only if BA = 0 a c b d = a + 4c b + 4d a c = 0 b d = 0 a + 4c = 0 b + 4d = One easily finds the general solution to this system is a = s, b = t, c = s, d = t. Thus A ker f if and only if s t 0 0 A = = s + t. s t Thus, {, } is a basis for ker f To find a basis for the image of f, first note that a matrix B V is in im f if and only if a c b d B = f(a) = a + 4c b + 4d for some a, b, c, d R. Thus, B im f if and only if B = a + b + c + d = (a c) + (b d) Thus, B im f if and only if B is in the span of T = {, }. Since 0 0 T is clearly linearly independent, T is a basis for im f.

5 7. The trace of a square matrix is defined [ to] be the sum of the entries on its main a b diagonal; e.g., the trace of the matrix is a + d. Now define f : M c d (R) R by f(a) = trace(a). (a) Show that f is a linear transformation. a b e g Let A = and B =. Then f(a + B) = trace(a + B) = (a + e) + c d h i (d + i) = (a + d) + (e + i) = trace(a) + trace(b) = f(a) + f(b). If c R then f(ca) = trace(ca) = (ca + cd) = c(a + d) = cf(a). (b) Find bases for the kernel and image of f. Let A be as in part (a). Then A ker f if and only if a + d = 0 if and only if a = d. Thus, A ker f if and only if a b A = c a = a + b + c Since the three matrices above span ker f and are clearly linearly independent, they form a basis for ker f. a 0 It is easy to see that f is onto. (If a R then f( ) = a.) A basis for im f 0 0 is therefore {1}. 8. Let f C(R) and define T : C(R) R by T (g) = 1 0 (a) Show that T is a linear transformation. fg dx. Let g 1, g C(R). Then T (g 1 + g ) = 1 f(g g ) dx = 1 fg 0 1 dx + 1 fg 0 dx = T (g 1 )+T (g ). Also, if c R then T (cg) = 1 cgf dx = c 1 gf dx = ct (g). Thus, 0 0 T is a linear transformation. (b) Prove that T is onto if and only if f(a) 0 for some a in the interval [0, 1]. (Hint: reason that T (f) > 0 if f(a) 0 for some a [0, 1].) One direction is easy: if T is onto then f(a) 0 for some a [0, 1]. For, if f(a) = 0 for all a [0, 1] then T (g) = 0 for all g C(R), contradicting that T is onto. For the converse, suppose f(a) 0 for some a [0, 1]. Then f(a) > 0 for some a [0, 1]. Since f is continuous, this means that the area between f and the x-axis over the interval [0, 1] is positive. Let c be this area. Then T (f) = 1 f dx = c > 0. To show T is onto, let a be any real number. Now, 0 a f C(R) (since c 0). As T is linear, we have T ( af) = at (f) = a c = a. c c c c This shows that every real number is in the image of T ; hence, T is onto.

6 (c) Prove that T is not one-to-one. Let g be any continuous function which is zero on the interval [0, 1] and non-zero for at least one value outside that interval. (There are many such functions.) Then g 0 (since 0 in this vector space is the function which is zero everywhere), but T (g) = 0. Thus, T is not one-to-one. 9. Let V = {a + bx + cy + dx + exy + fy a, b, c, d, e, f R} where x, y are variables. Then V is just the set of all polynomials in x and y of degree two or less. One can show that V is a vector space in much the same way as we showed P is a vector space. Now, consider the function T : V V by T (f) = f f. x y (a) Prove that T is a linear transformation. Let f, g V. Then T (f + g) = x (f + g) (f + g) y = x f + x g y f y g = ( x f y f) + ( x g y g) = T (f) + T (g). Similarly, one can show that T (cf) = ct (f) for any scalar c, using the fact that cf = c f. x x (b) Find bases for ker T and im T. First, we find a basis for ker T. Let f = a + bx + cy + dx + exy + fy V. Then f ker T if and only if f f = 0. Thus, f ker T if and only if x y b + dx + ey (c + ex + fy) = 0. Comparing coefficients, we get that f ker T if and only if b c = 0 d e = 0 e f = 0. The general solution to this system is a = s, b = t, c = t, d = u, e = u, f = u, where s, t, u are arbitrary real numbers. Therefore, f ker T if and only if for some s, t, u R we have f = s + tx + ty + ux + uxy + uy = s 1 + t(x + y) + u(x + xy + y ). Therefore, {1, x+y, x +xy +y } is a basis for ker T. (This set is clearly linearly independent and spans ker f by the above argument.) Thus, dim ker T = 3.

7 Next, we find a basis for im T. Let f = a + bx + cy + dx + exy + fy V. Then T (f) = b+dx+ey (c+ex+fy) = (b c)+(d e)x+(e f)y. Hence, im T is a subset of W = Span{1, x, y}. Is im T = W? This is true if and only if dim im T = dim W = 3. One way to show this is to use that dim im T + dim ker T = dim V. (We proved this in class on 3/13.) Since dim V = 6 and dim ker T = 3 (by above), we get dim im T = 3. A more straightforward way to prove this is to show 1, x, y im T. But 1 im T by letting a = e = d = f = c = 0 and b = 1; x im T by letting a = b = c = e = f = 0 and d = 1 ; and y im T by letting a = b = c = d = e = 0 and f = 1. Thus, im T = Span{1, x, y} and so {1, x, y} is a basis for im T. Yet another way to find a basis for the image of T is to use the fact that T (β) spans the image, where β is a basis for V. Thus, S = {T (1), T (x), T (y), T (x ), T (xy), T (y )} = {0, 1, 1, x, y x, y} spans im f. It should be clear that Span S = Span{1, x, y}. Therefore, {1, x, y} is a basis for im T.