Let $X$ be a finite dimensional vector space with the base $(e_1,…,e_n)$. Let $f\in X^*$. Then $$f(x)=\sum_{i=1}^n \lambda_i \beta_i,$$ for $x=\sum_{i=1}^{n}\lambda_ie_i$ and $\beta_{i}=f(e_i)$. We can see that if the base is given, then the funcional determine coefficients $\beta_i$ and conversly ($\star$). Given a base in $X$, I introduce $\phi\colon X^*\to \mathbb{R}$ by $$\phi(f)=(\beta_1,…,\beta_n)\in \mathbb{R}^{n}.$$ It is easy to see that $$\phi(f+g)=(\beta_1+\gamma_1,…,\beta_n+\gamma_{n})$$ and $$\phi(af)=(a\beta_1,…,a\beta_n).$$ What is more, $\phi$ is injection and it follows from $(\star)$. Why is $\phi$ surjective? I am not convinced that for any $(\beta_1,…,\beta_n)\in \mathbb{R}^{n}$ there exists $f\in X^*$ such that $\phi(f)=(\beta_1,…,\beta_n)$, because $\beta_i=f(e_i)$. What if there is no such a base? If $\phi$ is surjective, then it is isomorphism between $X^*$ and $\mathbb{R}^{n}$ which ends the proof.
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If any of the $\beta_i$ coefficients equal $0$ then $f(\mathbf x)$ is not surjective. – Doug M May 30 '17 at 20:29
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Please see this;https://math.stackexchange.com/questions/1128436/prove-that-vector-space-and-dual-space-have-same-dimension – xxxg Mar 17 '24 at 20:59
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Define the functionals $e_k^* \in X^*$ by $e_k^*(e_j) = \delta_{kj}$.
Then $\phi(\sum_k \beta_k e_k^*) = (\beta_1,\cdots, \beta_n)$. Hence $\phi$ is surjective.
In terms of the basis $e_k^*$ for $X^*$ and the basis $e_j$ for $X$ we see that $\phi$ is the identity matrix.
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