Linear Algebra - Covectors

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I need some help answering this question. I have answered part a: Suppose $\sum$ $a_ie^i=0$ for some $a_i \in K$. Then applying the vector $e_j$ to both sides we get $0=\sum a_i e^i(e_j)=\sum a_i \delta_{ij} = a^j, j=1,...n$. Hence the covectors $e^i$ are linearly independent. For part b I have: We have that $v=\sum a^i e_i$ then $T(v)= \sum a_i T(e_i)=\sum a_i$ Then I'm not sure what to do. I'm also not sure how I would approach part c and d.

Take finite-dimensional $V$.

Because $e_i$ is a basis for $V$, any vector in $v$ can be written $v=v^ie_i$. (Using Einstein summation here.) If T is any linear functional in $V^*$, this means $T(v)=v^iT(e_i)$ so the linear functional is specified completely if its action on the basis vectors $e_i$ is known.

Now, the covectors $e^i$ are nicely suited for telling us what the action on the basis vectors is. If we assume $T=t_je^j$, we can see that $T(e_i)=t_je^j(e_i)=t_j\delta^j_i=t_i$. This shows us that a $T$ of this form can be any element of $V^*$, as we can specify its action on all of the $e_i$ by specifying the coefficients $t_i$ (thus specifying its action on all vectors). So the covectors $e^i$ span $V^*$, and since you have already proven they are linearly independent, they are a basis.

The specific $T$ in the problem has all coefficients equal to $1$, of course.

For infinite dimensional $V$ we have to be more careful. Now a vector is $v=v^ie_i$ where only finitely many of the $v^i$ are non-zero. $T(v)$ is still specified by knowing all of the $T(e_i)$, but it is not necessarily the case that only finitely many of these are non-zero, so sums $t_ie^i$ with only finitely many non-zero $t_i$ do not span all of $V^*$. So they cannot be a basis.

The specific $T$ in the problem is an example of this: since $T(e_i)=1$ for all $e_i$, writing $T$ as a linear combination of the covectors $e^i$ would require an infinite number of terms, which is not permitted.

The fourth part is straightforward and you should be able to do it if you can follow this so far. :)