# Can this infinite vector still span a dual subspace?

Let's say I have a vector space $V$ of infinite dimension. Then, if we look at the dual basis, it does not actually span the whole dual vector space $V^*$, since linear combinations only allow finite sums. So if i take the canonical basis $B$ of $V$, then the subspace spanned by the dual basis does for example not contain a linear form $a^* : V\rightarrow K$ with $a^*(b_i) = 1$ for all $b_i$.

My question now is: If I took the subspace, which is spanned by the dual "basis" $b^*_i$ and swapped the first vector $b_1^*$ for the linear form $a^*$ I mentioned above, does this new family still span the subspace?

I removed one basis vector, but exchanged it for a linearly independent new vector. However, this new vector is not part of my subspace in the first place, so does it still span the subspace?

Any help is greatly appreciated!

The duals $b_i^*$ span a proper subspace if $V^*$, let's call that $W$. If you replace of of these $b_i^*$ with a vector $a^*\notin W$, then you get a different subspace $W'$ of $V^*$. After all, we clearly have $a^*\in W'\setminus W$ and $b_1^*\in W\setminus W'$. Nevertheless, $\{\,b_i^*\mid i\in I\,\}$ is a basis of $W$, and $\{\,b_i^*\mid i\in I, i\ne 1\,\}\cup \{a^*\}$ is a basis of $W'$. Just note that we still have no linear dependence due to nearly the same finiteness argument.
• Thank you very much! To me it just was not clear that $b_1^* \in W \setminus W'$, but I guess this immediately follows from the finiteness argument as you mentioned :) May 16 '17 at 20:17