Consider the vector-space $\mathbb{R}^3$. Now think of an arbitrary 1D-subspace of this vectorspace, lets say e.g. the subspace spanned by the basis vector $(1,1,0)^T$. This is a finite vector-space again – meaning it has a dual-basis with $b^{\ast}_i(b_j) = \delta_{ij}$, where $b^{\ast}_i$ is a basis of the dual-space and $b_j$ is the corresponding basis of the vector-space.
I always thought for a given basis the corresponding basis of the dual-space that fulfills the above property was unique. Now if we consider mentioned subspace, the dual-space basises that fulfill this property are (e.g. $(1,0,0), (0,1,0), (0.5,0.5,0)$ etc.) and therefore not unique. Where is my mistake?
Best Answer
You have to state things more carefully.
The dual basis is only unique if it's assumed to be contained in $V$. In the example you give, the relevant $V$ is the one-dimensional subspace. Of the three supposed dual bases you give, only one of them is contained in $V$.
Edit. That was assuming for convenience we have an inner-product space. Supposing $V$ is just a vector space, let $V'$ bbe the dual:
Why is the example you give not a counterexample? Here $V$ is a subspace of $\Bbb R^3$, as above. Now an element of $\Bbb R^3$ is not literally an element of $V'$. An element of $\Bbb R^3$ defines an element of $V'$ in a natural way, but the three vectors you mention all define the same element of $V'$. So you've only mentioned one element of $V'$, not contradicting uniqueness.