linear-guest

G.9 Linear Independence 405
notion of the dimension of a vector space to see why. So we think the vectors
v 1 , v 2 , v 3 and v 4 are linearly dependent, which means we need to show that there
is a solution to
α 1 v 1 + α 2 v 2 + α 3 v 3 + α 4 v 4 = 0
for the numbers α 1 , α 2 , α 3 and α 4 not all vanishing.
To find this solution we need to set up a linear system. Writing out the
above linear combination gives
4α 1 −3α 2 +5α 3 −α 4 = 0 ,
−α 1 +7α 2 +12α 3 +α 4 = 0 ,
3α 1 +4α 2 +17α 3 = 0 .
This can be easily handled using an augmented matrix whose columns are just
the vectors we started with
⎛
4 −3 5 −1
⎞
0 ,
⎝ −1 7 12 1 0 , ⎠ .
3 4 17 0 0 .
Since there are only zeros on the right hand column, we can drop it. Now we
perform row operations to achieve RREF
⎛
⎞ ⎛
⎞
71
4 −3 5 −1 1 0
25
− 4 25
⎝−1 7 12 1⎠ ⎜ 53 3 ⎟
∼ ⎝0 1
25 25⎠ .
3 4 17 0 0 0 0 0
This says that α 3 and α 4 are not pivot variable so are arbitrary, we set them
to µ and ν, respectively. Thus
(
α 1 = − 71
25 µ + 4 )
(
25 ν , α 2 = − 53
25 µ − 3 )
25 ν , α 3 = µ , α 4 = ν .
Thus we have found a relationship among our four vectors
(
− 71
25 µ + 4 ) (
25 ν v 1 + − 53
25 µ − 3 )
25 ν v 2 + µ v 3 + µ 4 v 4 = 0 .
In fact this is not just one relation, but infinitely many, for any choice of
µ, ν. The relationship quoted in the notes is just one of those choices.
Finally, since the vectors v 1 , v 2 , v 3 and v 4 are linearly dependent, we
can try to eliminate some of them. The pattern here is to keep the vectors
that correspond to columns with pivots. For example, setting µ = −1 (say) and
ν = 0 in the above allows us to solve for v 3 while µ = 0 and ν = −1 (say) gives
v 4 , explicitly we get
v 3 = 71
25 v 1 + 53
25 v 2 , v 4 = − 4 25 v 3 + 3 25 v 4 .
This eliminates v 3 and v 4 and leaves a pair of linearly independent vectors v 1
and v 2 .
405