A First Course in Linear Algebra, 2017a

202 R n ⃗u 1 =
Furthermore,
⎡
⎢
⎣
1
1
0
0
⎧⎡
⎪⎨
⎢
⎣
⎪⎩
⎤
⎥
⎦ ,⃗u 2 =
1
1
0
0
⎤
⎡
⎥
⎦ , ⎢
⎣
⎡
⎢
⎣
−1
0
1
0
−1
0
1
0
⎤
⎤ ⎡
⎥
⎦ , ⎢
⎣
⎥
⎦ ,⃗u 3 =
is linearly independent, as can be seen by taking the reduced row-echelon form of the matrix whose
columns are ⃗u 1 ,⃗u 2 and ⃗u 3 .
⎡
⎢
⎣
1 −1 1
1 0 0
0 1 0
0 0 1
⎤ ⎡
⎥
⎦ → ⎢
⎣
1
0
0
1
⎡
⎢
⎣
⎤⎫
⎪⎬
⎥
⎦
⎪⎭
1 0 0
0 1 0
0 0 1
0 0 0
Since every column of the reduced row-echelon form matrix has a leading one, the columns are
linearly independent.
Therefore {⃗u 1 ,⃗u 2 ,⃗u 3 } is linearly independent and spans V, soisabasisofV. Hence V has dimension
three.
♠
We continue by stating further properties of a set of vectors in R n .
Corollary 4.87: Linearly Independent and Spanning Sets in R n
The following properties hold in R n :
• Suppose {⃗u 1 ,···,⃗u n } is linearly independent. Then {⃗u 1 ,···,⃗u n } is a basis for R n .
• Suppose {⃗u 1 ,···,⃗u m } spans R n . Then m ≥ n.
•If{⃗u 1 ,···,⃗u n } spans R n , then {⃗u 1 ,···,⃗u n } is linearly independent.
⎤
⎥
⎦
1
0
0
1
⎤
⎥
⎦
Proof. Assume first that {⃗u 1 ,···,⃗u n } is linearly independent, and we need to show that this set spans R n .
To do so, let ⃗v be a vector of R n , and we need to write ⃗v as a linear combination of ⃗u i ’s. Consider the
matrix A having the vectors ⃗u i as columns:
A = [ ⃗u 1 ··· ⃗u n
]
By linear independence of the ⃗u i ’s, the reduced row-echelon form of A is the identity matrix. Therefore
the system A⃗x =⃗v has a (unique) solution, so⃗v is a linear combination of the ⃗u i ’s.
To establish the second claim, suppose that m < n. Then letting ⃗u i1 ,···,⃗u ik be the pivot columns of the
matrix [ ]
⃗u1 ··· ⃗u m