A First Course in Linear Algebra, 2017a

4.10. Spanning, Linear Independence and Basis in R n 193
The reduced row-echelon form is
⎡
⎢
⎣
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
and so every column is a pivot column and the corresponding system AX = 0 only has the trivial solution.
Therefore, these vectors are linearly independent and there is no way to obtain one of the vectors as a
linear combination of the others.
♠
Consider another example.
Example 4.68: Linear Independence
Determine whether the set of vectors given by
⎧⎡
⎤ ⎡ ⎤ ⎡
1 2
⎪⎨
⎢ 2
⎥
⎣ 3 ⎦
⎪⎩
, ⎢ 1
⎥
⎣ 0 ⎦ , ⎢
⎣
0 1
0
1
1
2
⎤
⎥
⎦
⎤ ⎡
⎥
⎦ , ⎢
⎣
is linearly independent. If it is linearly dependent, express one of the vectors as a linear combination
of the others.
3
2
2
−1
⎤⎫
⎪⎬
⎥
⎦
⎪⎭
Solution. Form the 4 × 4matrixA having these vectors as columns:
⎡
⎤
1 2 0 3
A = ⎢ 2 1 1 2
⎥
⎣ 3 0 1 2 ⎦
0 1 2 −1
Then by Theorem 4.66, the given set of vectors is linearly independent exactly if the system AX = 0has
only the trivial solution.
The augmented matrix for this system and corresponding reduced row-echelon form are given by
⎡
⎢
⎣
1 2 0 3 0
2 1 1 2 0
3 0 1 2 0
0 1 2 −1 0
⎤ ⎡
⎥
⎦ →···→ ⎢
⎣
1 0 0 1 0
0 1 0 1 0
0 0 1 −1 0
0 0 0 0 0
Not all the columns of the coefficient matrix are pivot columns and so the vectors are not linearly independent.
In this case, we say the vectors are linearly dependent.
It follows that there are infinitely many solutions to AX = 0, one of which is
⎡
⎢
⎣
1
1
−1
−1
⎤
⎥
⎦
⎤
⎥
⎦