00:01
So we want to prove that a set of vectors s is linearly independent.
00:08
If and only if, every finite subset of vectors is also linearly independent.
00:24
So let's put a vectors is also linearly independent.
00:34
Now note that s is not necessarily finite.
00:37
So we can't just trivially assume the implication this way, because s is a subset of itself, but if it's not finite, you run into trouble.
00:48
So it will be easier to do both of these with proof by contradiction.
00:53
So first, for the forward direction, suppose that s is linearly independent, but there exists a finite subset.
01:17
That is not.
01:23
And let's call that finite subset v1 through vk.
01:30
So therefore, and this again, is a subset of s.
01:39
Therefore, as it is not linearly independent, there exist constants not all zero, c i such that c1 v1 plus c2 v2 all the way to ck minus 1 vk equals k vk by the supposition that this is not linearly independent...