Show that $\delta$ functions are in dual space Suppose $\{x_1,\dots,x_n\}$ is a basis for vector space $X$
Show that there $\exists$ linear function $\{e_1,\dots,e_n\}$ in dual space $X'$ such that $e_i(x_j)=\delta_{i,j}=\cases{1 & \text{if } i=j, \\ 0 & \text{if }  i\neq j.}$
But I have been told not to use the canonical form of basis, (i.e. if $z$ is an i-th basis vector, you can not write that it's represented as a vector with only one non-zero component).
And we also "don't know" that delta functions form a basis in dual space.
I know how to show that using canonical form of basis vector, but I am stuck without using it, could you please show how to solve it this way?
 A: Hint: Show that for any tuple of numbers $a_1,\dots a_n$, the map $\sum c_i x_i \mapsto \sum a_i c_i$ is linear and hence in $X'$. This should tell you which maps to use as $e_i$, and how they form a basis of $X'$.
A: Since $x_1,\ldots,x_n$ is a basis for $X$, we know that for any $x \in X$ there are unique scalars $e_1(x),\ldots,e_n(x)$ such that 
$$ x = e_1(x) \cdot x_1 + \ldots e_n(x) \cdot x_n.$$
Doing this for every $x \in X$, we get scalar valued functions $e_1,\ldots,e_n$ on $X$. The question remains, are they linear? Suppose $z = ax + by$ where $a,b$ are scalars and $x,y \in X$. We know
\begin{align*}
x = e_1(x)x_1 + \ldots e_n(x) x_n && y = e_1(y) x_1 + \ldots e_n(y) x_n
\end{align*}
so, adding and gathering terms, we also have
$$ z = ax+by = (ae_1(x) + be_1(y)) x_1 + \ldots (ae_n(x) + be_n(y)) x_n.$$ 
But on the other hand, 
$$z= e_1(z)x_1 + \ldots e_n(z) x_n$$
where the scalars in front of the $x_i$ are supposedly unique. So, we get that $e_i(z) = ae_i(x)  + be_i(y)$ for $i=1,\ldots,n$. So, $e_1,\ldots,e_n$ are linear functionals. It remains to check that $e_i(x_j) = \delta_{i,j}$.  I will leave this point for you to consider.
