# Linear independence of covectors

Let $\ V$ be a vector space over a field $\ \Bbb K$. Let $\ e_i, i \in I$ be a basis for $\ V$, such that any element can be written uniquely as $\ \sum_J a_j e_j$ where $J \subset I$ is a finite subset and $a_j \in \Bbb K \setminus {\{0}\}$ for all $j \in J$. The dual vector space $\ V^{*}$ is defined as the set of all linear maps $\ V \to \Bbb K$. Elements of this space are covectors. Given the basis of V, covectors $\ e^{i} \in V^{*}, i \in I$ are defined by $\ e^{i}(e_{j}) = \delta_{ij} (1$ if $\ i=j$, 0 otherwise).

How would I go about showing that the covectors are linearly independent? Further, how would I show that they are spanning (and thus form a basis), assuming $V$ is finite dimensional?

Suppose $\sum a_ie^i = 0$. What is $\sum a_ie^i(e_j)$?
Let $f \in V^\ast$. Then $f(\sum a_ie_i) = \sum a_if(e_i)$. Can you figure out a sum of $e^i$ that has the same values?
• For the first, this equals $\sum a_{i} \delta_{ij}$ which is equals $a^{j}$? I don't quite follow - is this sufficient to show linear independence? For the second part, would it be equal to $\sum f_i e^{j}(e_{i}) = \sum f_i \delta_{ij} = f_{i}$, where each $f_{i}$ is some scalar coefficient? Then each one can be chosen so spans? – Ian Baker Nov 1 '15 at 21:07
• On the first, note that "$=0$" part of the equation. On the second, you need to pick the $f_i$ to match the given functional $f$. If you show that $f(v) = \sum f_ie^i(v)$ for all $v \in V$, then you know that $f = \sum f_ie^i$, and since $f$ is arbitrary, that means the $e^i$ span $V^\ast$. – Paul Sinclair Nov 1 '15 at 22:20