MATH 223, Linear Algebra
Fall, 2010
Solutions to Assignment 3
1. Let
V
=
M
2
(
R
) be the real vector space of 2
×
2 matrices with real entries.
For each of the following subsets of
V
, decide if it is independent, if it is
a spanning set for
V
, and/or if it is a basis for
V
. Justify your answers.
(a)
‰±
1
3
2

1
¶
,
±
2 0
5 0
¶
,
±
4
1
0

1
¶²
.
Solution: This set is independent. To see this, suppose the matrices
in it are
A
,
B
and
C
and that
α
1
A
+
α
2
B
+
α
3
C
= 0. Then
α
1
+
2
α
2
+ 4
α
3
= 0, 3
α
1
+ 0
α
2
+ 1
α
3
= 0, 2
α
1
+ 5
α
2
+ 0
α
3
= 0 and

1
α
1
+ 0
α
2

1
α
3
= 0. We could easily enough set this set of four
equations in three unknowns as some kind of matrix (4
×
3, in fact)
but here it’s easier to note that the last equation forces
α
3
=

α
1
and the second forces
α
3
=

3
α
1
; together these force
α
1
=
α
3
= 0
and then
α
2
= 0 follows from any of the other two. This is the point
— only the trivial linear combination comes out to be zero.
M
2
(
R
) is 4d.; this set has only three vectors (i.e. 2
×
2 matrices), so
it cannot be a spanning set for
V
. It is therefore not a basis for
V
.
(b)
‰±
1 1
1 1
¶
,
±
1
1

1

1
¶
,
±
1

1
1

1
¶
,
±
1 2
3 5
¶²
.
Solution: This has just the right number of elements to be a basis for
V
, so we need only check whether it is independent. Suppose that
the matrices are
A
,
B
,
C
and
D
(in order); if
aA
+
bB
+
cC
+
dD
= 0,
this implies that
a
+
b
+
c
+
d
= 0,
a
+
b

c
+2
d
= 0,
a

b
+
c
+3
d
= 0
and
a

b

c
+ 5
d
= 0. It is not hard to see that this system has
only the trivial solution
a
=
b
=
c
=
d
= 0; if necessary rowreduce
the matrix
1
1
1 1
1
1

1 2
1

1
1 3
1

1

1 5
. The set is independent and thus a
basis for
M
2
(
R
).
(c)
‰±
1 0
0 0
¶
,
±
1 1
0 0
¶
,
±
1 1
1 0
¶
,
±
1 1
1 1
¶
,
±
0 1
1 1
¶²
.
Solution: Because there are ﬁve matrices here, the set cannot be
independent, and so it isn’t a basis for
M
2
(
R
). It is, however, a
spanning set for
V
. Suppose that
A
=
±
a
b
c d
¶
is any vector in
V
;
it is not hard to see that (for instance)
(
a

b
)
±
1 0
0 0
¶
+(
b

c
)
±
1 1
0 0
¶
+(
c

d
)
±
1 1
1 0
¶
+
d
±
1 1
1 1
¶
=
1