A First Course in Linear Algebra, 2017a

194 R n
Therefore we can write
⎡
1⎢
⎣
1
2
3
0
⎤ ⎡
⎥
⎦ + 1 ⎢
⎣
2
1
0
1
⎤ ⎡
⎥
⎦ − 1 ⎢
⎣
0
1
1
2
⎤ ⎡
⎥
⎦ − 1 ⎢
⎣
3
2
2
−1
⎤ ⎡
⎥
⎦ = ⎢
⎣
0
0
0
0
⎤
⎥
⎦
This can be rearranged as follows
⎡ ⎤ ⎡
1
1⎢
2
⎥
⎣ 3 ⎦ + 1 ⎢
⎣
0
2
1
0
1
⎤
⎡
0
⎥
⎦ − 1 ⎢ 1
⎣
1
2
⎤
⎡
⎥
⎦ = ⎢
This gives the last vector as a linear combination of the first three vectors.
Notice that we could rearrange this equation to write any of the four vectors as a linear combination of
the other three.
♠
When given a linearly independent set of vectors, we can determine if related sets are linearly independent.
Example 4.69: Related Sets of Vectors
Let {⃗u,⃗v,⃗w} be an independent set of R n .Is{⃗u +⃗v,2⃗u + ⃗w,⃗v − 5⃗w} linearly independent?
⎣
3
2
2
−1
⎤
⎥
⎦
Solution. Suppose a(⃗u +⃗v)+b(2⃗u +⃗w)+c(⃗v − 5⃗w)=⃗0 n for some a,b,c ∈ R. Then
Since {⃗u,⃗v,⃗w} is independent,
(a + 2b)⃗u +(a + c)⃗v +(b − 5c)⃗w =⃗0 n .
a + 2b = 0
a + c = 0
b − 5c = 0
This system of three equations in three variables has the unique solution a = b = c = 0. Therefore,
{⃗u +⃗v,2⃗u + ⃗w,⃗v − 5⃗w} is independent.
♠
The following corollary follows from the fact that if the augmented matrix of a homogeneous system
of linear equations has more columns than rows, the system has infinitely many solutions.
Corollary 4.70: Linear Dependence in R n
Let {⃗u 1 ,···,⃗u k } be a set of vectors in R n . If k > n, then the set is linearly dependent (i.e. NOT
linearly independent).
Proof. Form the n × k matrix A having the vectors {⃗u 1 ,···,⃗u k } as its columns and suppose k > n. ThenA
has rank r ≤ n < k, so the system AX = 0 has a nontrivial solution and thus not linearly independent by
Theorem 4.66.
♠