Page 206 - 35Linear Algebra

206 Linear Independence

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i. First, we show that if v k = c v 1 + · · · c k−1 v k−1 then the set is linearly
dependent.

This is easy. We just rewrite the assumption:

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c v 1 + · · · + c k−1 v k−1 − v k + 0v k+1 + · · · + 0v n = 0.

This is a vanishing linear combination of the vectors {v 1 , . . . , v n } with
not all coefficients equal to zero, so {v 1 , . . . , v n } is a linearly dependent
set.

ii. Now we show that linear dependence implies that there exists k for
which v k is a linear combination of the vectors {v 1 , . . . , v k−1 }.

The assumption says that

1
2
n
c v 1 + c v 2 + · · · + c v n = 0.
Take k to be the largest number for which c k is not equal to zero. So:

1
2
k
c v 1 + c v 2 + · · · + c k−1 v k−1 + c v k = 0.
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(Note that k > 1, since otherwise we would have c v 1 = 0 ⇒ v 1 = 0,
contradicting the assumption that none of the v i are the zero vector.)

So we can rearrange the equation:

2
1
k
c v 1 + c v 2 + · · · + c k−1 v k−1 = −c v k
c 1 c 2 c k−1
⇒ − v 1 − v 2 − · · · − v k−1 = v k .
c k c k c k
Therefore we have expressed v k as a linear combination of the previous
vectors, and we are done.

Worked proof

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