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 transforma-tion T : P2 → 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 P2 if and only if

T (β) =

110

,

101

,

011

is a basis for R3. (This follows from Theorem 8 on page 244. For further expla-

nation, see Exercises 25 and 26 on page 249.) To check T (β) is a basis for R3,

it suffices to put the three columns in 3 × 3 matrix and show that the rref of thismatrix is the identity matrix. (This computation is trivial, so I won’t reproduceit 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 + c2 = 7

c1 + c3 = −5

c2 + c3 = 3.

Solving this system, we get c1 = −1

2, c2 = 15

2, c3 = −9

2. Thus, the coordinate

vector of 7 − 5x + 3x2 with respect to β is

−1

215

2

−9

2

.

2. Let S = {1 − x2 + 3x3,−1 + x + 2x2 − x3,−3 + 5x + 8x2 + x3, 4x + 5x2 + 3x3} be asubset 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)), whereT (S) is the set of coordinate vectors of the vectors in S:

T (S) =

10−13

,

−112−1

,

−3581

,

0453

.

To find a basis for the span of T (S), we write these vectors as the columns of amatrix and find its rref (which I did using Maple):

1 −1 −3 00 1 5 4−1 2 8 53 −1 1 3

→ · · · →

1 0 2 00 1 5 00 0 0 10 0 0 0

.

From the rref, we see that our original matrix has pivots in the first, second, andfourth columns. Therefore, the first, second, and fourth polynomials form a basisfor 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 thecoordinate 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 thatc1(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 whoseaugmented matrix is:

1 −1 0 −90 1 4 35−1 2 5 483 −1 3 23

which reduces to

1 0 0 100 1 0 190 0 1 40 0 0 0

.

Since the system is consistent, f is in the span of S; its coordinate vector with

respect to the basis we found in part (a) is

10194

.

3. Let f : U → V be a linear transformation of vector spaces. Prove that im f is asubspace 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). Thismeans 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 provesim 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, . . . ) thenai = 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 TS = I?

Yes, since (TS)((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 invertibil-ity and right invertibility of linear transformations of infinite dimensional vectorspaces?

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 arelinear transformations from V to V where V is a finite dimensional vector spaceand TS = I, then ST = I also. (This follows from what we have proved aboutsquare matrices.)

5. Let S = {2 sin2 x, 3 cos2 x, cos 2x} be a subset of C(R), the vector space of continuousfunctions 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 c1, c2, c3

such that c1x+ 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 c1x+ c2 sin x = 0. Letting x = π, we seethat c1 = 0. This leaves us with c2 sin x = 0. Letting x = π

2, we obtain c2 = 0.

6. Let B =

[

1 −2−2 4

]

. Let V = M2×2(R), the vector space of 2 × 2 matrices with real

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 havef(cA) = B(cA) = c(BA) = cf(A). Thus, f is a linear transformation.

(b) Find a bases for the kernel and image of f .

We first find a basis for ker f . Let A =

[

a b

c d

]

. Then

A ∈ ker f ⇐⇒ f(A) = 0

⇐⇒ BA = 0

⇐⇒

[

a − 2c b − 2d−2a + 4c −2b + 4d

]

=

[

0 00 0

]

.

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

A =

[

2s 2ts t

]

= s

[

2 01 0

]

+ t

[

0 20 1

]

.

Thus, {

[

2 01 0

]

,

[

0 20 1

]

} 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 ifand only if

B = f(A) =

[

a − 2c b − 2d−2a + 4c −2b + 4d

]

for some a, b, c, d ∈ R. Thus, B ∈ im f if and only if

B = a

[

1 0−2 0

]

+ b

[

0 10 −2

]

+ c

[

−2 04 0

]

+ d

[

0 −20 4

]

= (a − 2c)

[

1 0−2 0

]

+ (b − 2d)

[

0 10 −2

]

.

Thus, B ∈ im f if and only if B is in the span of T = {

[

1 0−2 0

]

,

[

0 10 −2

]

}. Since

T is clearly linearly independent, T is a basis for im f .

7. The trace of a square matrix is defined to be the sum of the entries on its main

diagonal; e.g., the trace of the matrix

[

a b

c d

]

is a + d. Now define f : M2×2(R) → R

by f(A) = trace(A).

(a) Show that f is a linear transformation.

Let A =

[

a b

c d

]

and B =

[

e g

h i

]

. Then f(A + B) = trace(A + B) = (a + e) +

(d + i) = (a + d) + (e + i) = trace(A) + trace(B) = f(A) + f(B). If c ∈ R thenf(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 ifa = −d. Thus, A ∈ ker f if and only if

A =

[

a b

c −a

]

= a

[

1 00 −1

]

+ b

[

0 10 0

]

+ c

[

0 01 0

]

.

Since the three matrices above span ker f and are clearly linearly independent,they form a basis for ker f .

It is easy to see that f is onto. (If a ∈ R then f(

[

a 00 0

]

) = a.) A basis for im f

is therefore {1}.

8. Let f ∈ C(R) and define T : C(R) → R by T (g) =∫

1

0fg dx.

(a) Show that T is a linear transformation.

Let g1, g2 ∈ C(R). Then T (g1 + g2) =∫

1

0f(g1 + g2) dx =

∫

1

0fg1 dx +

∫

1

0fg2 dx =

T (g1)+T (g2). Also, if c ∈ R then T (cg) =∫

1

0cgf dx = c

∫

1

0gf 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, iff(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 > 0for some a ∈ [0, 1]. Since f is continuous, this means that the area betweenf 2 and the x-axis over the interval [0, 1] is positive. Let c be this area. Then

T (f) =∫

1

0f 2 dx = c > 0. To show T is onto, let a be any real number. Now,

acf ∈ C(R) (since c 6= 0). As T is linear, we have T ( a

cf) = a

cT (f) = a

c· c = a.

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-zerofor 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 + fy2 | 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 canshow that V is a vector space in much the same way as we showed P2 is a vector space.Now, consider the function T : V → V by T (f) = ∂

∂xf − ∂

∂yf .

(a) Prove that T is a linear transformation.

Let f, g ∈ V . Then

T (f + g) =∂

∂x(f + g) −

∂

∂y(f + g)

=∂

∂xf +

∂

∂xg −

∂

∂yf −

∂

∂yg

= (∂

∂xf −

∂

∂yf) + (

∂

∂xg −

∂

∂yg)

= T (f) + T (g).

Similarly, one can show that T (cf) = cT (f) for any scalar c, using the fact that∂∂x

cf = c ∂∂x

f .

(b) Find bases for ker T and im T .

First, we find a basis for ker T . Let f = a + bx + cy + dx2 + exy + fy2 ∈ V .Then f ∈ ker T if and only if ∂

∂xf − ∂

∂yf = 0. Thus, f ∈ ker T if and only if

b + 2dx + ey − (c + ex + 2fy) = 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 forsome s, t, u ∈ R we have

f = s + tx + ty + ux2 + 2uxy + uy2

= s · 1 + t(x + y) + u(x2 + 2xy + y2).

Therefore, {1, x+y, x2 +2xy+y2} is a basis for ker T . (This set is clearly linearlyindependent 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 + fy2 ∈ V . ThenT (f) = b+2dx+ey−(c+ex+2fy) = (b−c)+(2d−e)x+(e−2f)y. Hence, im T is asubset 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 (byabove), we get dim im T = 3. A more straightforward way to prove this is to show1, 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

2; and y ∈ im T by letting

a = b = c = d = e = 0 and f = − 1

2. 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 (y2)}

= {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} isa basis for im T .