If I have a vector space and a dual space , with a basis of and a basis of , does the basis of always have to have this form:
?
It's easy to prove that this is a basis, but i'm wondering if the basis can only be like this.
If it is, how can I prove it?
No, I don' think so.
Proof: We have this theorem that any set of independednt vectors can be expanded to find the basis of a finite-dim vector space. So start with a linear operator which is not in the above format. This can then be expanded to include other linear operators to form the basis of a dual space. And this bases will not be the form you suggested (as the staring operator in not in that form)
[Edit] After seeing Tonio's post. I would rather issue a warning for my post. It might not be correct. But I think as long as a 'fixed' basis of (for the context of this question) I should still be right.
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!)?
Yes - That's exactly the argument I was trying to give. But as Tonio said, given , we can always find a basis of U where these linear operators operate on that basis of U as per your original form.
@Tonio-It would be great if you can plz validate our understanding here.