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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!

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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.

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  • $\begingroup$ 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 :) $\endgroup$
    – Jack4t3
    May 16 '17 at 20:17

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