TITLE: Why is the Kronecker delta a base for the dual vector space?
QUESTION [5 upvotes]: The Kronecker delta can be defined like this:
$\delta_{ij} = \begin{cases}
1, & \text{if } i = j   \\
0, & \text{if } i \ne j \end{cases}$
The dual space is a vector space $V^*$ that can be defined to any $\mathbb{K}-$vector space $V$. Its elements are linear transformations $\Phi: V \rightarrow \mathbb{K}$.
According to my script and Wikipedia a base to the dual space can be found quite easily. If $X = \{x_i\}_{i = 1, 2, ..., n}$ is a base of V, then $X^* = \{x_i^*\}_{i = 1, 2, ...,n}$ with $x_i^*(x_j) = \delta_{ij}$.
Now I chose the $\mathbb{R}^1$-vector space with $\mathbb{K} = \mathbb{R}$ to get a really simple example. Then $X = \{1\}$ and $X^* = \{x_1^*\}$ with $x_1^* (x) = 
\begin{cases}
1, & \text{if } x = 1 \\
0, & \text{if } x \neq 1
\end{cases}$. So this should be a base for some linear transformations. But all linear transformations $\Phi : V \rightarrow W$ have to fulfil two conditions:

$\forall x, y \in V: \Phi(x+y) = \Phi(x) + \Phi(y)$
$\forall a \in \mathbb{K}: \forall x \in V: \Phi(a \cdot x) = a \cdot \Phi(x)$

But $x_1^*(1 + -1) = x_1^*(0) = 0 \neq 1 = 1 + 0 = x_1^*(1) + x_1^*(-1)$. So $x_i^*$ is no element of the dual space but its base?!? Can you please tell me where I got something wrong?

REPLY [2 votes]: You should think of $x^*_i$ as the $i$-th coordinate function for the basis $X$, in other words the map that associates to any vector $v$ the coefficient of $x_i$ when $v$ is expressed on the basis $X$. You may write this maps as
$$
  x^*_i: v=v_1x_1+\cdots+v_nx_n\mapsto v_i
$$
Now it should be clear that when you take $v=x_j$ then you find that $x^*_i(x_j)=1$ if $j=i$, and  $x^*_i(x_j)=0$ otherwise. That is more succinctly expressed by $x_i^*(x_j) = \delta_{i,j}$.