A First Course in Linear Algebra, 2017a

472 Vector Spaces
Proof. Suppose ∑ k i=1 c i⃗u i + d⃗v =⃗0. It is required to verify that each c i = 0andthatd = 0. But if d ≠ 0,
then you can solve for⃗v as a linear combination of the vectors, {⃗u 1 ,···,⃗u k },
⃗v = −
k
∑
i=1
( ci
)
⃗u i
d
contrary to the assumption that⃗v is not in the span of the ⃗u i . Therefore, d = 0. But then ∑ k i=1 c i⃗u i =⃗0 and
the linear independence of {⃗u 1 ,···,⃗u k } implies each c i = 0also.
♠
Consider the following example.
Example 9.24: Adding to a Linearly Independent Set
Let S ⊆ M 22 be a linearly independent set given by
{[ ] [ 1 0 0 1
S = ,
0 0 0 0
Show that the set R ⊆ M 22 given by
{[ 1 0
R =
0 0
is also linearly independent.
] [ 0 1
,
0 0
]}
] [ 0 0
,
1 0
]}
Solution. Instead of writing a linear combination of the matrices which equals 0 and showing that the
coefficients must equal 0, we can instead use Lemma 9.23.
To do so, we show that [ 0 0
1 0
]
{[ 1 0
/∈ span
0 0
] [ 0 1
,
0 0
]}
Write
[
0 0
1 0
]
[
1 0
= a
0 0
[ ]
a 0
=
0 0
[ ]
a b
=
0 0
]
+
[
0 1
+ b
0 0
[ ]
0 b
0 0
]
Clearly there are no possible a,b to make this equation true. Hence the new matrix does not lie in the
span of the matrices in S. By Lemma 9.23, R is also linearly independent.
♠