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Let $X$ be an n-dimensional normed space, where $n\in\mathbb{N}$. Show that the space $L(X,\mathbb{K})$ of bounded linear functionals on $X$ is al n-dimensional.

Suppose $\{x_1,x_2,...,x_n\}$ is a basis for $X$, define $f_k:X\rightarrow\mathbb{K}$ by $f_k(\Sigma_{j=1}^{n}a_jx_j)=a_k$. Then $\{f_1,f_2,...,f_n\}$ forms a basis for the space of linear functionals on $X$.

How do I actually show that the basis $\{f_1,f_2,...,f_n\}$ spans the space of all linear functionals on $X$?

Any help would be appreciated. Thanks!

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Here's a hint: Given $f\in L(X,\mathbb{K})$, for $k=1,\ldots,n$ put $\beta_k=f(x_k)$. Then show that $f=\sum_{k=1}^n\beta_kf_k$, by showing that $f(x)=\left(\sum_{k=1}^n\beta_kf_k\right)(x)$ for all $x\in X$.

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Let $g \colon X \to \mathbb K$ be a linear functional on $X$. This is a linear map, and a linear map is uniquely determined by the images of the basis vectors of $X$, namely $b_1 = g(x_1), \dots, b_n = g(x_n)$. Now let's try to write $g$ as a linear combination of $f_1, \dots f_n$. If we try $g_0 = b_1 f_1 + b_2 f_2 + \cdots + b_n f_n$, and look at where it sends the basis vectors $x_1, \dots, x_n$, we can see that $g_0(x_1) = b_1 f_1(x_1) + 0 + \dots + 0 = b_1$, $g_0(x_2) = 0 + b_2 f_2(x_2) + 0 + \dots + 0 = b_2$, and so on until $g_0(x_n) = 0 + \cdots + 0 + b_n f_n(x_n) = b_n$. This agrees with the definition of $g$, thus $g$ is a linear combination of $f_1, \dots, f_n$.

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