Post on 27-Mar-2018
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 .