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This is from Kreyszig's Introductory Functional Analysis Theorem 2.9-1.

Let $X$ be an n-dimensional vector space and $E=\{e_1, \cdots, e_n \}$ a basis for $X$. Then $F = \{f_1, \cdots, f_n\}$ given by (6) is a basis for the algebraic dual $X^*$ of $X$, and $\text{dim}X^* = \text{dim}X=n$.

Other background info:

$f_k(e_j)=\delta_{jk} = 0 \text{ if } j \not = k, 1 \text{ if } j=k$ (Kronecker delta).

Proof. $F$ is a linearly independent set since

\[\sum_{k=1}^n \beta_k f_k (x) = 0\]

with $x = e_j$ gives

\[\sum_{k=1}^n \beta_k f_k (e_j) = \sum_{k=1}^n \beta_k \delta_{jk} = \beta_j = 0\]

so that all the $\beta_k$'s are zero, as we go through each $e_j$.

And it continues to show that $F$ spans $X^*$.So we show that when we use some specific $x$'s, namely $x=e_j$, then $\beta_k = 0$. But, don't we want to show that $\beta_k$'s are all zero for any $x \in X$? Why does showing that the scalars are zero one at a time, when applying the basis elements of $X$ suffice?

I think I'm missing something really simple here..

Let $X$ be an n-dimensional vector space and $E=\{e_1, \cdots, e_n \}$ a basis for $X$. Then $F = \{f_1, \cdots, f_n\}$ given by (6) is a basis for the algebraic dual $X^*$ of $X$, and $\text{dim}X^* = \text{dim}X=n$.

Other background info:

$f_k(e_j)=\delta_{jk} = 0 \text{ if } j \not = k, 1 \text{ if } j=k$ (Kronecker delta).

Proof. $F$ is a linearly independent set since

\[\sum_{k=1}^n \beta_k f_k (x) = 0\]

with $x = e_j$ gives

\[\sum_{k=1}^n \beta_k f_k (e_j) = \sum_{k=1}^n \beta_k \delta_{jk} = \beta_j = 0\]

so that all the $\beta_k$'s are zero, as we go through each $e_j$.

And it continues to show that $F$ spans $X^*$.So we show that when we use some specific $x$'s, namely $x=e_j$, then $\beta_k = 0$. But, don't we want to show that $\beta_k$'s are all zero for any $x \in X$? Why does showing that the scalars are zero one at a time, when applying the basis elements of $X$ suffice?

I think I'm missing something really simple here..

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