Bases of a covector space

Question

I want to prove this proposition.

Let $ V $ be a finite dimensional vector space. Given any basis $ E_1,E_2,...,E_n $ for V, let $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n \in V^* $ be the covectors defined by

$$ \varepsilon^i(E_j)= \delta_j^i $$

Then $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n $ is a basis for $ V^* $ and $ dim V^* = dim V$

I have started the proof as follows.

First to prove that the covectors $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n $ are linearly independent.

If $ \sum_{i=0}^n r_i\varepsilon^i = 0 $, then

$ r_1\varepsilon^1 + r_2\varepsilon^2 + ... + r_n\varepsilon^n (E_1) = 0 (E_1)= 0 $

$r_1\varepsilon^1 (E_1)=0 $

$ r_1=0 $

Similarly each $r_i =0 $ which implies the covectors $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n $ are linearly independent .

Now i am struck at proving that $\varepsilon^1,\varepsilon^2,...,\varepsilon^n $ span $ V^* $

The fact that $ dim V = dim V^* $ is not yet known.

Answer 1

For the last part do \begin{eqnarray*} F(x)&=&(f-\sum_{i=1}^nf(E_i)\varepsilon^i)(x),\\ &=&f(x)-\sum_{i=1}^nf(E_i)\varepsilon^i(x), \end{eqnarray*} then for $x=E_k$ you get \begin{eqnarray*} F(E_k)&=&f(E_k)-\sum_{i=1}^nf(E_i)\varepsilon^i(E_k),\\ &=&f(E_k)-\sum_{i=1}^nf(E_i)\delta^i_k,\\ &=&f(E_k)-f(E_k),\\ &=&0. \end{eqnarray*} So $F\equiv0$ and then $$f=\sum_{i=1}^nf(E_i)\varepsilon^i.$$

Answer 2

Note that an element $f \in V^*$ is uniquely determined by its values on any basis of $V$.

Now compare $f$ to $\sum_{i = 1}^n f(E_i) \varepsilon^i$ on $E_1,\dots,E_n$ and conclude that these elements of $V^*$ coincide.