Linear Algebra, 2020a

114 Chapter Two. Vector Spaces
⎛
⎜
1
⎞ ⎛
⎟ ⎜
−3
⎞
⎛ ⎞ ⎛
1
⎟
⎜ ⎟ ⎜
0
⎞
⎟
dependent: { ⎝0⎠ , ⎝ 0 ⎠} independent: { ⎝0⎠ , ⎝1⎠}
0 0
0 0
We got the dependent superset by adding a vector from the x-axis and so the
span did not grow. We got the independent superset by adding a vector that
isn’t in [S], because it has a nonzero y component, causing the span to grow.
For the independent set
⎛ ⎞ ⎛
1
⎜ ⎟ ⎜
0
⎞
⎟
S = { ⎝0⎠ , ⎝1⎠}
0 0
the span [S] is the xy-plane. Here are two supersets.
⎛
⎜
1
⎞ ⎛
⎟ ⎜
0
⎞ ⎛ ⎞
⎛ ⎞ ⎛
3
1
⎟ ⎜ ⎟
⎜ ⎟ ⎜
0
⎞ ⎛
⎟ ⎜
0
⎞
⎟
dependent: { ⎝0⎠ , ⎝1⎠ , ⎝−2⎠} independent: { ⎝0⎠ , ⎝1⎠ , ⎝0⎠}
0 0 0
0 0 1
As above, the additional member of the dependent superset comes from [S],
the xy-plane, while the added member of the independent superset comes from
outside of that span.
Finally, consider this independent set
⎛
⎜
1
⎞ ⎛
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
0
⎞
⎟
S = { ⎝0⎠ , ⎝1⎠ , ⎝0⎠}
0 0 1
with [S] =R 3 . We can get a linearly dependent superset.
⎛
⎜
1
⎞ ⎛
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
0
⎞ ⎛ ⎞
2
⎟ ⎜ ⎟
dependent: { ⎝0⎠ , ⎝1⎠ , ⎝0⎠ , ⎝−1⎠}
0 0 1 3
But there is no linearly independent superset of S. One way to see that is to
note that for any vector that we would add to S, the equation
⎛ ⎞ ⎛
x
⎜ ⎟ ⎜
1
⎞ ⎛
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
0
⎞
⎟
⎝y⎠ = c 1 ⎝0⎠ + c 2 ⎝1⎠ + c 3 ⎝0⎠
z 0 0 1
has a solution c 1 = x, c 2 = y, and c 3 = z. Another way to see it is that we
cannot add any vectors from outside of the span [S] because that span is R 3 .