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We have to solve the following:

Let $V$ be an $n$-dimensional vector space over $\mathbb{K}$ and let $B=(v_{1}, ... ,v_{n})$ be the basis of $V$. Every $x \in V$ can be described as $$x=\sum_{k=1}^{n}c_{k}v_{k}$$ with $c_{k} \in \mathbb{K}$.

  1. Let $j \in \mathbb{N}: 1\leq j\leq n$. Show that the map $w_{j}: V\rightarrow \mathbb{K}$, which transforms $x$ into $c_j$, is a linear functional!
  2. Show that $(w_{1}, ... ,w_n)$ is a basis of the dual space $V^*$.
  3. Show that the linear map $V\rightarrow V^*$ is an isomorphism.

What I did:

  1. I don't understand the first question at all...
  2. I would say that it's the basis of $V^*$ because $\dim(V)=\dim(V^*)$, meaning the basis must have $n$ elements.
  3. I showed that it's bijective, is that enough?

I'm really stuck with this problem, it would be very nice if someone could help!

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1) For prove this you just have to apply the definition of a linear functional for every $w_j$, i.e $\forall$ x, y $\in V$ $w_j(x + y)=w_j(x)+w_j(y)$ and $\forall$ $\alpha \in K$ $w_j(\alpha x)=\alpha w_j(x)$

I) Let $x=\sum x_i v_i$ and $x=\sum y_i v_i$ elements of $V$ and $j \in \{1, ... , n\} $, then

$w_j(x+y)=w_j(\sum x_i v_i + \sum y_i v_i) = w_j(\sum (x_i + y_i) v_i)=x_j + y_j = w_j(x)+w_j(y)$

II) Let $\alpha \in K$ and $x$ as before

$w_j(\alpha x)=w_j(\alpha \sum x_i v_i)=w_j(\sum \alpha x_i v_i)=\alpha x_j = \alpha w_j(x)$

2) For this we just have to show that $\{w_1, ... , w_n\}$ generate $V$ (the part of linear independence is free from the dimension as you say in your comments), so let $f \in V^*$ a linear functional and $x \in V$, then

$f(x)=f(\sum x_iv_i)=\sum x_if(v_i)=\sum w_j(x)f(v_i)$ i.e $f$ is a linear combination of $\{w_1, ... , w_n\}$ (the first time that you see this could be weird, but give it some time and you will got it!).

3) Define the function $\phi:V \rightarrow V^* $ such that $\sum x_i v_i \mapsto \sum x_iw_i$ (the good part of this function is that send basis into basis, so you can use this for prove the isomorphism easily).

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  1. Each vector $x$ in $V$, once you chose a basis, can be represented as a $n$-uple $(c_1,\dots,c_n) \in \mathbb{K}^n$. You can consider the map: $w_j: V \to \mathbb{K}$ defined by $w_j(v) = c_j$, i.e. the map that gives you the $j$-th coordinate. You have to show this is linear.

  2. In order to show that $\{w_1,\dots,w_n\}$ is a basis you have to show that each $f \in V^*$ can be written as $\sum_{j=1}^n \alpha_j w_j$

  3. You have to show that the map $V \to V^*$ is bijective, linear and also the inverse is linear

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