### Basis

Exercise Prove that a vector space is finite dimensional ⇔ The space can be spanned by finitely many vectors.

Answer An earlier exercise suggests that if a finitely set `B` of vectors spans `V`, then by deleting some vectors from `B`, we may find a basis of `V`. In particular, the dimension of `V` is the number of vectors in a subset of `B`, so that `V` is finite dimensional.

So assume `B` = {`b`_{1}, `b`_{2}, ... , **b**_{k}} is a finite set of vectors that spans `V`. Then we ask whether `B` is linearly independent. If `B` is not linearly independent, then one vector, say **b**_{k}, is a linear combination of the others. By this property, the subset {`b`_{1}, `b`_{2}, ... , **b**_{k-1}} still spans `V`. Then we ask again whether {`b`_{1}, `b`_{2}, ... , **b**_{k-1}} is linearly independent. If not, we can delete a vector, say **b**_{k-1}, so that the rest {`b`_{1}, `b`_{2}, ... , **b**_{k-2}} still spans `V`. We continue asking the same question until the answer is yes. Then we get a linearly independent subset of B, say {`b`_{1}, `b`_{2}, ... , **b**_{n}}, that still spans `V`. The subset {`b`_{1}, `b`_{2}, ... , **b**_{n}} is then a basis of `V`.

Remark A basis of `V` must span `V` and must be linearly independent. In case the spanning condition if satisfied, we may deleting "wasted" vectors one by one, until there is no more waste. Since the deleted vectors are "wasted", the span remains to be `V`. Since there is no more waste in the end, we get a linearly independence after deletion. This is the key idea in the argument above. This exercise gives a concrete example on the process.

Remark Since a basis can be obtained by deleting vectors from a spanning set, we have this numerical relation for a spanning set as a consequence. Similarly, a basis can be obtained by adding vectors to a linearly independent set (see this exercise for concrete examples). as a result, we have this numerical relation for a linearly independent set.