Solution

Clearly, the three vectors span a subspace of [Maple Math] . Let us call it [Maple Math] . We first claim that [Maple Math] . Clearly, [Maple Math] contains [Maple Math] and since [Maple Math] , we see that [Maple Math] is also in [Maple Math] . Since a spanning set of [Maple Math] is contained in [Maple Math] , we must have all of [Maple Math] contained in [Maple Math] . Since [Maple Math] is already a subspace of [Maple Math] , the two are equal!

Now, if a spanning set of a vector space is not a basis, then it must be dependent and we can pick a basis by discarding some subset. Thus, if [Maple Math] were not a basis of [Maple Math] , then some subset of 1 or 2 elements will be a basis, but this is a contradiction to [Maple Math] having dimension [Maple Math] (or a basis of three elements)!

So, [Maple Math] must be a basis of [Maple Math] .

More generally, we can argue that [Maple Math] together with [Maple Math] is a basis, as well as [Maple Math] together with [Maple Math] is a basis.

More generally, if [Maple Math] is a basis of some vector space [Maple Math] and if [Maple Math] is a linear combination, then it can be swapped in the basis for any [Maple Math] if its coefficient [Maple Math] is nonzero! This exchange principle is the heart of the theory of abstract vector spaces!