Math 314H Solutions to Homework # 1 1. Let β = {1 + x, 1 + x2 , x + x2 } be a subset of P2 . (a) Prove that β is a basis for P2 . Let δ = {1, x, x2 } be the standard basis for P2 and consider the linear transformation T : P2 → R3 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 P2 if and only if        1 0   1      1 , 0 , 1 T (β) =   0 1 1

is a basis for R3 . (This follows from Theorem 8 on page 244. For further explanation, see Exercises 25 and 26 on page 249.) 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 + 3x2 with respect to β. We need to find scalars c1 , c2 , c3 such that c1 (1 + x) + c2 (1 + x2 ) + c3 (x + x2 ) = 7 − 5x + 3x2 . Comparing coefficients, this leads to the system c1 + c 2 = 7 c1 + c3 = −5 c2 + c3 = 3. Solving this system, we get c1 = − 12 , c2 = vector of 7 − 5x + 3x2 with respect to β is  1 −2  15  . 2 − 92

15 , 2

c3 = − 29 . Thus, the coordinate

2. Let S = {1 − x2 + 3x3 , −1 + x + 2x2 − x3 , −3 + 5x + 8x2 + x3 , 4x + 5x2 + 3x3 } be a subset of P3 . (a) Find a basis for Span(S). As in Problem 1, we use the coordinate mapping T : P3 → R4 defined by T (f ) = [f ]δ , where δ = {1, x, x2 , x3 }. 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:         1 −1 −3 0             0 1 5  ,   ,   , 4  . T (S) =  −1  2   8  5      3 −1 1 3

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):     1 −1 −3 0 1 0 2 0 0   1 5 4   → · · · → 0 1 5 0 . −1 2 0 0 0 1 8 5 3 −1 1 3 0 0 0 0 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 − x2 + 3x3 , −1 + x + 2x2 − x3 , 4x + 5x2 + 3x3 }. (b) Decide whether f = −9 + 35x + 48x2 + 23x3 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 c1 , c2 , c3 such that c1 (1−x2 +3x3 )+c2 (−1+x+2x2 −x3 )+c3 (4x+5x2 +3x3 ) = −9+35x+48x2 +23x3 . Comparing coefficients as we did in Problem 1, we get a system of equations whose augmented matrix is:     1 0 0 10 1 −1 0 −9 0 1 0 19 0 1 4 35      which reduces to 0 0 1 4  . −1 2 5 48  0 0 0 0 3 −1 3 23 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 19. 4 3. 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 v1 , v2 be vectors in im f . Then v1 = f (u1 ) and v2 = f (u2 ) for some u1 , u2 ∈ U . Then v1 + v2 = f (u1 ) + f (u2 ) = f (u1 + u2 ). This means that v1 + v2 is in im f , since v1 + v2 = f (w), where w = u1 + u2 . Similarly, let c ∈ R be a scalar. Then cv1 = cf (u1 ) = f (cu1 ). Thus, cv1 ∈ 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 ((a0 , a1 , a2 , . . . )) = (a1 , a2 , a3 , . . . ), and S((a0 , a1 , a2 , . . . )) = (0, a0 , a1 , a2 , . . . ) for all sequences (a0 , a1 , a2 , . . . ) in R∞ . Furthermore, let I : R∞ → R∞ be the identity map; i.e., I((a0 , a1 , a2 , . . . )) = (a0 , a1 , a2 . . . ).

(a) Find the image and kernel of T . Since T ((0, a0 , a1 , a2 , . . . )) = (a0 , a1 , a2 , . . . ), we see that every vector in R∞ is in the image of T ; i.e., im T = R∞ . Suppose (a0 , a1 , . . . ) ∈ ker T . Then T (a0 , a1 , . . . ) = (a1 , a2 , . . . ) = (0, 0, 0, . . . ). Therefore, ai = 0 for all i ≥ 1. Hence, ker T = {(a0 , 0, 0, 0, · · · | a0 ∈ R}. (b) Find the image and kernel of S. Clearly, ker S = {(0, 0, 0, . . . )}, since if S((a0 , a1 , a2 , . . . )) = (0, 0, 0, . . . ) then ai = 0 for all i. Also, im T = {(0, a1 , a2 , a3 , . . . ) | ai ∈ 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)((a0 , a1 , a2 , . . . )) = T (S((a0 , a1 , a2 , . . . ))) = T ((0, a0 , a1 , . . . )) = (a0 , a1 , a2 , . . . ). (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 = {2 sin2 x, 3 cos2 x, cos 2x} 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 2x = cos2 x−sin2 x, we get 3(2 sin2 x)−2(3 cos2 x)+6(cos 2x) = 0. Hence, S is linearly dependent. The second set is linearly independent. To prove this, assume we have scalars c 1 , c2 , c3 such that c1 x + c2 sin x + c3 cos x = 0. This equation must hold for all x. Letting x = 0, we have c3 = 0. Thus, the equation reduces to c1 x + c2 sin x = 0. Letting x = π, we see that c1 = 0. This leaves us with c2 sin x = 0. Letting x = π2 , we obtain c2 = 0. 

 1 −2 6. Let B = . Let V = M2×2 (R), the vector space of 2 × 2 matrices with real −2 4 entries. Define f : V → V by f (A) = BA.

(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 ⇐⇒ BA = 0     0 0 a − 2c b − 2d . = ⇐⇒ 0 0 −2a + 4c −2b + 4d Thus, A ∈ ker f if and only if a − 2c = 0 b − 2d = 0 −2a + 4c = 0 −2b + 4d = 0. One easily finds the general solution to this system is a = 2s, b = 2t, c = s, d = t. Thus A ∈ ker f if and only if       2s 2t 2 0 0 2 A= =s +t . s t 1 0 0 1     0 2 2 0 } is a basis for ker f . , Thus, { 0 1 1 0 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 − 2c b − 2d B = f (A) = −2a + 4c −2b + 4d for some a, b, c, d ∈ R. Thus, B ∈ im f if and only if         0 −2 −2 0 0 1 1 0 +d +c +b B=a 0 4 4 0 0 −2 −2 0     1 0 0 1 = (a − 2c) + (b − 2d) . −2 0 0 −2     1 0 0 1 Thus, B ∈ im f if and only if B is in the span of T = { , }. Since −2 0 0 −2 T is clearly linearly independent, T is a basis for im f .

7. The trace of a square matrix is defined  tobe the sum of the entries on its main a b is a + d. Now define f : M2×2 (R) → R diagonal; e.g., the trace of the matrix c d 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       1 0 0 1 0 0 =a +b +c . 0 −1 0 0 1 0 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) =

R1 0

f g dx.

(a) Show that T is a linear transformation. R1 R1 R1 Let g1 , g2 ∈ C(R). Then T (g1 + g2 ) = 0 f (g1 + g2 ) dx = 0 f g1 dx + 0 f g2 dx = R1 R1 T (g1 ) + T (g2 ). Also, if c ∈ R then T (cg) = 0 cgf dx = c 0 gf dx = cT (g). Thus, T is a linear transformation. (b) Prove that T is onto if and only if f (a) 6= 0 for some a in the interval [0, 1]. (Hint: reason that T (f ) > 0 if f (a) 6= 0 for some a ∈ [0, 1].) One direction is easy: if T is onto then f (a) 6= 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) 6= 0 for some a ∈ [0, 1]. Then f (a) 2 > 0 for some a ∈ [0, 1]. Since f is continuous, this means that the area between f 2 and the x-axis over the interval [0, 1] is positive. Let c be this area. Then R1 T (f ) = 0 f 2 dx = c > 0. To show T is onto, let a be any real number. Now, a f ∈ C(R) (since c 6= 0). As T is linear, we have T ( ac f ) = ac T (f ) = ac · c = a. c This shows that every real number is in the image of T ; hence, T is onto.

(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 6= 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 + dx2 + exy + f y 2 | 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 P2 is a vector space. ∂ ∂ f − ∂y f. Now, consider the function T : V → V by T (f ) = ∂x (a) Prove that T is a linear transformation. Let f, g ∈ V . Then ∂ ∂ (f + g) − (f + g) ∂x ∂y ∂ ∂ ∂ ∂ = f+ g− f− g ∂x ∂x ∂y ∂y ∂ ∂ ∂ ∂ f) + ( g − g) =( f− ∂x ∂y ∂x ∂y = T (f ) + T (g).

T (f + g) =

Similarly, one can show that T (cf ) = cT (f ) for any scalar c, using the fact that ∂ ∂ cf = c ∂x f. ∂x (b) Find bases for ker T and im T . First, we find a basis for ker T . Let f = a + bx + cy + dx2 + exy + f y 2 ∈ V . ∂ ∂ Then f ∈ ker T if and only if ∂x f − ∂y f = 0. Thus, f ∈ ker T if and only if b + 2dx + ey − (c + ex + 2f y) = 0. Comparing coefficients, we get that f ∈ ker T if and only if b−c=0 2d − e = 0 e − 2f = 0. The general solution to this system is a = s, b = t, c = t, d = u, e = 2u, 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 + ux2 + 2uxy + uy 2 = s · 1 + t(x + y) + u(x2 + 2xy + y 2 ). Therefore, {1, x + y, x2 + 2xy + y 2 } 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.

Next, we find a basis for im T . Let f = a + bx + cy + dx2 + exy + f y 2 ∈ V . Then T (f ) = b+2dx+ey−(c+ex+2f y) = (b−c)+(2d−e)x+(e−2f )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 = 21 ; and y ∈ im T by letting a = b = c = d = e = 0 and f = − 12 . 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 (x2), T (xy), T (y 2)} = {0, 1, −1, 2x, y − x, 2y} spans im f . It should be clear that Span S = Span{1, x, y}. Therefore, {1, x, y} is a basis for im T .