Download 4.3.1) Yes, it is a subspace. It is clearly a subset of R2

4.3.1) Yes, it is a subspace. It is clearly a subset of R2 . To show that it is
a subspace, Theorem 4.3 says that we need only check that it is closed under
addition and scalar multiplication. For the former if (x, x) and (y, y) are two
arbitrary points on the line, then their sum, (x, x) + (y, y) = (x + y, x + y),
also has both coordinates the same, so it is also on the line. For the latter, for
arbitrary c ∈ R and (x, x) on the line, the scalar multiple c(x, x) = (cx, cx) also
has both coordinates the same, so it is on the line.
1
4.3.4) No, it is not a subspace. The point
is in W , but its scalar
1
1
2
multiple 2
=
is not.
1
2
4.3.6) a) subspace. The sum of any two vectors with last coordinate zero
also has last coordinate zero. So does any scalar multiple of any vector with
last coordinate zero.




1
1
b) not a subspace.  1  is in the set, but its scalar multiple (−1)  1  =
1
1


−1
 −1  is not.
−1
c) subspace. The sum of any two vectors with the first two coordinates the
same also has the first two coordinates the same. So does any scalar multiple
of any vector with the first two coordinates the same.
 


0
0
d) not a subspace.  0  is in the set, but its scalar multiple 2  0  =
1
1


0
 0  is not.
2
a b
e f
and B =
be two
c d
g h
arbitrary matrices
in W , and let k be an arbitrary scalar. We can compute the
a+e b+f
sum A + B =
, and the sum of the entries of this matrix is
c+g d+h
a+e+b+f +c+g+d+h = (a+b+c+d)+(e+f +g+h)= 0+0 = 0,
so A+B ∈ W .
ka kb
Likewise, we can compute the scalar multiple kA =
, which has the
kc kd
sum of its coordinates as ka + kb + kc + kd = k(a + b + c + d) = k(0) = 0, so
kA ∈ W .
4.3.13) Yes, it is a subspace. Let A =
1
4.4.1) It was probably easiest to do this problem by giving the standard basis,
and then tweaking it in some way that leaves the span obviously equivalent to
the original for the second spanning set.
a)

 
 1
 0 ,

0

 
 2
 0 ,

0
 
0
1 ,
0
 
0
2 ,
0

0 
0 

1

0 
0 

2
b)
1
0
1
0
0 1
0 0
0 0
,
,
,
0 0
1 0
0 1
0
0 2
0 0
0 0
,
,
,
0
0 0
3 0
0 4
0
0
c)
{1, t, t2 , t3 }
{0, 1, t, t2 , t3 }
4.4.2) a) 1 is not in the span of the set, as all vectors in the set have a
constant term of 0.
0
b)
is not in the span of the set, as all vectors in the set have a second
1
coordinate of 0.
1 0
c)
is not in the span of the set, as both vectors in the set have the
0 0
top left and bottom right coordinates the same.
4.4.5) (Not real solutions this time, but just answers.) b does not span R2 ,
but the others do.
4.4.10) (Again, not real solutions, but just the answer.) Yes, the set does
span M22 .
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4.5.22) If v1 , v2 , and v3 are linearly dependent, then there are constants c1 ,
c2 , and c3 , not all zero, such that c1 v1 + c2 v2 + c3 v3 = 0. We can compute
(c1 − c3 )w1 + (c3 − c2 )w2 + c2 w3
=
(c1 − c3 )v1 + (c3 − c2 )(v1 + v3 ) + c2 (v1 + v2 + v3 )
= c1 v1 − c3 v1 + c3 v1 − c2 v1 + c3 v3 − c2 v3 + c2 v1 + c2 v2 + c2 v3
= c1 v1 + c2 v2 + c3 v3
= 0.
If (c1 − c3 ), (c3 − c2 ), and c2 are all zero, then c2 = 0, from which c3 = 0, and
so c1 = 0. Since c1 , c2 , and c3 are not all zero, (c1 − c3 ), (c3 − c2 ), and c2 are
not all zero, and so we have a nontrivial linear combination of w1 , w2 , and w3
which gives 0. Therefore, T is linearly dependent.
Alternatively, this problem is easier if we borrow the notion of dimension
from section 4.6. Let W = span S. By Theorem 4.9, some subset of S is a
basis for W . It cannot be all of S, as S is linearly independent, and hence not
a basis. Therefore, W has a basis consisting of at most two vectors in S, and
so dim W ≤ 2. Each vector in T is in span S, so span T ⊆ W . Therefore, the
span of T has dimension at most two. By Corollary 4.4, a set of three vectors
in a vector space of dimension at most two cannot be linearly independent.
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