umm, what do you mean by "dual to another basis in V"?

Also aman_cc, does a linear operator exist which is not in that form?

sends a vector to a scalar. I'll choose the scalar product and see if that gives me a basis.

ie.

This is linearly independent (since it's the only vector we have in our prospective basis) so we can extend it:

Basis:

(and it doesn't matter what

is

since all we want is a basis where at least one basis vector is not in the form outlined in my previous post).

We have a basis, but we know that

is not in the form in my previous post. Hence we have found a basis for

that is not in the form in my previous post.

Is this how the argument goes (but better worded of course!)?