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Math 110 Week 2 Homework Solutions
1
Section 1.4
1.
(a) True. It’s obtained by setting every coefficient equal to zero. It is even, by definition,
the unique linear combination of the empty set of vectors.
(b) False. By definition, the span of ∅ is {0}, the zero subspace.
(c) True. By Theorem 1.5, Span(S) is contained in any subspace of V that contains S.
Thus, Span(S) is contained in the intersection of all such subspaces. In the other direction, note that Span(S) is a subspace of V which contains S, and so it contains
the aforementioned intersection. Thus, Span(S) both contains and is contained in the
intersection of all subspaces of V which contain S, and so they are equal.
(d) False. If we multiply an equation by 0, we lose it.
(e) True. This operation is reversible, and hence it is allowed. To see that it is reversible,
note that we can return to the original configuration by instead subtracting the same
multiple of the first equation from the (modified) second one.
(f) True. The zero vector is a solution to any system of linear equations.
2
(b) We first subtract the third equation from the first to obtain
x1 − 6x2 + 6x3 = 4
x1 − 2x2 + x3 = 3
2x1 − x2 − 2x3 = 6
Then we subtract the first equation from the second and twice the first equation from
the third
x1 − 6x2 + 6x3 = 4
4x2 − 5x3 = −1
11x2 − 14x3 = −2
We subtract thrice the second from the third
x1 − 6x2 + 6x3 = 4
4x2 − 5x3 = −1
−x2 + x3 = 1
Math 110 Week 2 Homework Solutions
2
the negate the third and exchange it with the second
x1 − 6x2 + 6x3 = 4
x2 − x3 = −1
4x2 − 5x3 = −1
We add multiples of the second equation to the others to eliminate all appearances of
x2 .
x1
= −2
x2 − x3 = −1
−x3 = 3
Finally, we negate the third equation and add it to the second.
x1
= −2
x2 = −4
x3 = −3
This gives (x1 , x2 , x3 ) = (= 2, −4, −3).
(d) We first subtract the first equation from the second and third to get
x1 +
2x2 + 2x3
= 2
−2x2 + 6x3 + 5x4 = −8
−x2 + 3x3 + 5x4 = 1
We now negate the third equation and swap it with the second
x1 +
2x2 + 2x3
= 2
x2 − 3x3 − 5x4 = −1
−2x2 + 6x3 + 5x4 = −8
We add the third equatio to the first, and then we add twice the second equation to the
third
x1
+ 8x3 + 5x4 = −6
x2 − 3x3 − 5x4 = −1
−5x4 = −10
Finally, we use the third equation to remove the x4 terms, and then we divide it by −5.
x1
+ 8x3
x2 − 3x3
= −16
=
9
x4 =
2
This gives (x1 , X2 , x3 , x4 ) = (−8x3 − 16, 3x3 + 9, x3 , 2).
Math 110 Week 2 Homework Solutions
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(f) We begin by swapping the top two equations with the bottom two
3x1 + x2 − x3
x1 + 3x2 + 10x3
x1 + 2x2 + 6x3
2x1 + x2 + x3
= 15
= −5
= −1
= 8
Next, we subtract the third from the second and the fourth from the first
− 2x3
x2 + 4x3
x1 + 2x2 + 6x3
2x1 + x2 + x3
x1
= 7
= −4
= −1
= 8
Now we use the first equation to remove the remaining x1 terms
x1
− 2x3
x2 + 4x3
2x2 + 8x3
x2 + 5x3
= 7
= −4
= −8
= −6
Using the second equation to remove the other x2 terms and swapping the last two
equations gives
x1
− 2x3 = 7
x2 + 4x3 = −4
x3 = −2
0= 0
Finally, we use the third equation to remove the x3 terms from the first two
x1
= 3
x2 = 4
x3 = −2
0= 0
This gives (x1 , x2 , x3 ) = (3, 4, −2).
3
(b) We want to solve
−3x1 + 2x2 = 1
2x1 − x2 = 2
x1 − x2 = −3
Math 110 Week 2 Homework Solutions
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Using the third equation to remove the other x1 terms, we get
−x2 = −8
x2 = 8
x1 − x2 = −3
From here, we see that (x1 , x2 ) = (5, 8) is a solution, so that (1, 2, −3) = 5(−3, 2, 1) +
8(2, −1, −1). In other words, the first vector is indeed a linear combination of the other
two.
(d) We want to solve
x 1 + x2 = 2
2x1 − 3x2 = −1
−3x1 + 2x2 = 0
Using the first equation to remove the other x1 terms, we get
x1 +
x2 = 2
−5x2 = −5
5x2 = 6
The last two equations can be added to show 1 = 0, which means that there is no
solution. In other words, the first vector is not a linear combination of the other two.
(f) We want to solve
x1 − 3x2 = −2
2x1 − 3x2 = 2
−x1 + 3x2 = 2
Using the first equation to remove the other x1 terms, we get
x1 − 3x2 = −2
3x2 = 6
0= 0
From here, we see that (x1 , x2 ) = (4, 2) is a solution, so that (−2, 2, 2) = 4(1, 2, −1) +
2(−3, −3, 3). In other words, the first vector is indeed a linear combination of the other
two.
Math 110 Week 2 Homework Solutions
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(b) If so, then (−1, 2, 1) = x1 (1, 0, 2) + x2 (−1, 1, 1). From the second component, we see
that x2 must be 2. In other words, (−1, 2, 1) = x1 (1, 0, 2) + (−2, 2, 2), or (1, 0, −1) =
x1 (1, 0, 2), which is impossible. So the given vector is not in the span of S.
(d) We see by examining the first two components that if so, then it must be that
(2, −1, 1, −3) = 2(1, 0, 1, −1) − (0, 1, 1, 1).
We see that this is in fact true, so the given vector is in the span of S.
(f) If so, then 2x3 −x2 +x+3 = a(x3 +x2 +x+1)+b(x2 +x+1)+c(x+1). Looking just at the x
and constant components, we see that we would need x+3 = a(x+1)+b(x+1)+c(x+1).
This is impossible, since the coefficients of x and 1 are equal on the right hand side but
different on the left. Thus, the first vector is not in the span of S.
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Let (a, b, c) ∈ F3 . If F has characteristic not equal to 2, then
(a, b, c) =
a−b+c
−a + b + c
a+b−c
(1, 1, 0) +
(1, 0, 1) +
(0, 1, 1).
2
2
2
So (a, b, c) is in the span of the given vectors, and so they generate F3 . If F has characteristic
2, then the claim is actually false.
8
Let p ∈ Pn (F ). The p = an xn + · · · + a0 x0 , which expresses p as a linear combination of
elements of {1, x, . . . , xn }. So that set generates Pn (F ).
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a b
Let M be a symmetric 2 × 2 matrix. Then M =
, so that M = aM1 + cM2 + bM3 .
b c
Thus, M is in the span of {M1 , M2 , M3 }, and so the given matrices generate the space of
symmetric 2 × 2 matrices.
16
Suppose v is in the span of S, and that v = b1 u1 + · · · + bn uk and v = c1 w1 + · · · + cm wm ,
where bi , c1 ∈ F and ui , wi ∈ S. Let v = {u1 , . . . , uk } ∪ {w1 , . . . , wm } = {v1 , . . . , vn }. Set
(
bj
if vi = uj
ai =
0
otherwise
Math 110 Week 2 Homework Solutions
and
(
cj
a0i =
0
6
if vi = wj
otherwise
Then v = a1 v1 + · · · + an vn = a01 v1 + · · · + a0n vn , so that (a1 − a01 )v1 + · · · + (an − a0n )vn = 0.
This implies ai − a0i for all i, so that the two linear combinations were the same.
Section 1.5
3

 
 
 
 

1 1
0 0
0 0
1 0
0 1
0 0 + 1 1 + 0 0 − 1 0 − 0 1 = 0
0 0
0 0
1 1
1 0
0 1
5
Suppose an xn + · · · + a0 x0 is the zero polynomial. Then all its coefficients are zero, and so
the given set is linearly independent.
9
First, we show that if {u, v} is linearly dependent, then one is a multiple of the other. For
this, note that we have au + bv = 0 where either a 6= 0 or b 6= 0 (or both). If a 6= 0, then
u = − ab v, so that u is a multiple of v. If not, then b 6= 0, so that v = − ab u says that v is a
multiple of u.
Next, we show the other direction, so suppose one of u or v is a multiple of the other. If
u = cv, then 1u + (−c)v = 0 has a nonzero coefficient for u, so {u, v} is linearly dependent.
If not, then v = cu and cu + (−1)v = 0 has a nonzero coefficient for v, so {u, v} is linearly
dependent.
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{(1, 0, 0), (0, 1, 0), (1, 1, 0)}.
Section 1.6
1.
(a) False. The empty set is a basis for the zero vector space.
(b) True. This follows from Theorem 1.9.
(c) False. The vector space P(F ) of polynomials of arbitrary degree over F does not have
a finite basis.
Math 110 Week 2 Homework Solutions
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(d) False. {1} and {2} are different bases for R.
(e) True. This is Corollary 1.
(f) False. Pn (F ) has dimension n + 1.
(g) False. Mm×n (F ) has dimension mn.
(h) True. Let n be the dimension of V . Then Corollary 2(c) implies that |S1 | ≤ n, while
the Replacement Theorem implies n ≤ |S2 |.
(i) False. {1, 2} generates R, but 0 can be written both as 0 · 1 + 0 · 2 and as (−2) · 1 + 1 · 2.
(j) True. This follows from Theorem 1.11.
(k) True. {0} is the only subspace of dimension 0, while V itself is the only subspace of
dimension n.
(l) True. This follows by combining (a) and (b) in Corollary 2.
3.
(b) By Corollary 2(b), it is enough to check whether the given set is linearly independent.
In other words, we need to see whether a(1 + 2x + x2 ) + b(3 + x2 ) + c(x + x2 ) = 0 implies
a = b = c = 0. To do this, we attempt to solve the system of equations
a + 3b
=0
2a
+c=0
a+ b+c=0
Eliminating the a terms from the second and third row, we get
a+
3b
=0
−6b + c = 0
−2b + c = 0
Subtracting thrice the third equation from the second gives
a+
3b
=0
−2c = 0
−2b + c = 0
From here, we see that c = 0, so b = 0, so a = 0. Thus, the set is linearly independent,
and hence it is a basis.
Math 110 Week 2 Homework Solutions
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(d) The process is identical to the previous problem. Here, the system of equations is
−a + 3b − 2c = 0
2a − 4b − 5c = 0
4a − 10b − 6c = 0
Eliminating the a terms from the second and third row, we get
−a + 3b − 2c = 0
2b − 9c = 0
2b − 14c = 0
Subtracting the second equation from the third gives
−a + 3b −
2b −
2c = 0
9c = 0
−5c = 0
From here, we see that c = 0, so b = 0, so a = 0. Thus, the set is linearly independent,
and hence it is a basis.
4.
No. P3 (R) has dimension 4, while there are only three given polynomials. By Corollary
2(a), the polynomials cannot generate P3 (R).
5.
No. R3 has dimension 3, so any basis has three elements, while the given set has four
elements. By Corollary 2(c), a set with more than three elements cannot be linearly independent, since it would be impossible to extend it to a basis.
8.
Note first that the set {(1, −1, 0, 0, 0), (0, 1, −1, 0, 0), (0, 0, 1, −1, 0), (0, 0, 0, 1, −1)} is linearly
independent and sits in W , so W must have dimension at least 4. On the other hand,
(1, 0, 0, 0, 0) 6∈ W , so W can’t be all of R5 , and hence dim(W ) < 5. So W is 4-dimensional.
Thus, what we are looking for is a linearly independent set of four ui .
To do so, start by noting that {u6 , u7 } is linearly independent as witnessed by the zeros in
the first two coordinates. Next, note that if u1 were in the span of {u6 , u7 }, then from the
first two coordinates we would have u1 = u6 + 2u7 = (2, −3, −22, 15, 8), which is false. Thus,
u1 is not in the span of {u1 , u2 }, and so {u1 , u6 , u7 } is linearly independent. We need to add
one more vector to obtain a basis.
Math 110 Week 2 Homework Solutions
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For this, there are faster tricks where one replaces {u1 , u6 , u7 } with {u1 − u6 − 2u7 , −1
u , u7 },
3 6
but the most straightforward approach is trial and error. One immediately sees that u2 =
−3u1 , but after setting up and solving a system of equations, u3 ends up working. Thus, a
basis is {u1 , u3 , u6 , u7 }.
11.
Since {u, v} has is a basis with two elements, we see that V is two-dimensional. Thus,
checking whether a given two-element subset S ⊂ V is a basis amounts to checking whether
it generates V . Since V = Span({u, v}), Theorem 1.5 implies that it is enough to check
whether u and v are in Span(S), since then
Span({u, v}) ⊂ Span(S) ⊂ V
so they are all equal.
·(au).
For S = {u+v, au}, we use that a 6= 0, so we can write u = a1 ·(au) and v = 1·(u+v)+ −1
a
So S is a basis for V .
For S = {au, bv}, we use that a, b 6= 0, so we can write u = a1 · (au) and v = 1b · (bv). So S is
a basis for V .
12.
The process here is identical to the previous problem, with two replaced by three. Thus, we
need only check that {u, v, w} ⊂ Span({u + v + w, v + w, w}). For this, we write
u = 1 · (u + v + w) − 1 · (v + w)
v = 1 · (v + w) − 1 · (w)
w = 1 · (w)
So {u + v + w, v + w, w} is a basis for V .
14.
For W1 , we note that
{(1, 0, −1, 0, 0), (1, 0, 0, −1, 0), (0, 1, 0, 0, 0), (0, 0, 0, 0, 1)} ⊂ W1
is linearly independent, so W has dimension at least 4. On the other hand, (1, 0, 0, 0, 0) 6∈ W1 ,
so W1 6= F5 , so dim(W1 ) < 5. Thus, W1 is 4-dimensional and
{(1, 0, −1, 0, 0), (1, 0, 0, −1, 0), (0, 1, 0, 0, 0), (0, 0, 0, 0, 1)}
is a basis for it.
For W2 , we observe that the equations imply that an element of F5 belongs to W2 if and
only if it is of the form (a1 , a2 , a2 , a2 , −a1 ), or eqivalently a1 (1, 0, 0, 0, −1) + a2 (0, 1, 1, 1, 0).
Thus, W2 is 2-dimensional and a basis is {(1, 0, 0, 0, −1), (0, 1, 1, 1, 0)}.
Math 110 Week 2 Homework Solutions
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15.
Let E ij be as in Example 3 of Section 1.6. Define B ij for (i, j) 6= (n, n) by
(
E ij
if i 6= j
B ij =
ij
i+1 j+1
E −E
if i = j
Then
B = {B ij : 1 ≤ i ≤ n, 1 ≤ j ≤ n, (i, j) 6= (n, n)}
is a linearly independent subset of W with n2 − 1 elements. Since E 11 ∈
6 W , we have
W 6= Mn×n (F ), so that dim(W ) < n2 . Thus, W must have dimension n2 − 1 and B is a
basis for it.
26.
Let W = {f ∈ Pn (R) : f (a) = 0} be the given subset. Note that
{x − a, x2 − a2 , . . . , xn − an } ⊂ W
is linearly independent, so W has dimension at least n. On the other hand, 1 6∈ W , so that
dim(W ) < dim(Pn (R)) = n + 1. Thus, W must have dimension exactly n.