Originally Posted by

**Showcase_22** 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?

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

ie. $\displaystyle e^i (e_j)=<e_j,e_j>=||e_j||^2$

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

Basis: $\displaystyle \{ e^i, f^1,.....,f^{n-1} \}$ (and it doesn't matter what $\displaystyle f^p$ is $\displaystyle p=1,...,n-1$ 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 $\displaystyle e^i$ is not in the form in my previous post. Hence we have found a basis for $\displaystyle V*$ that is not in the form in my previous post.

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