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Lemma Let $V$ is a finite-dimensional vector space (over the filed $F$), say $n=\dim V$. If $x_1,\ldots,x_n$ is a basis of $V$, then there exists a basis $f_1,\ldots,f_n$ of the dual space $V'$ such that $f_i(x_j)=\delta_{ij}$, $i,j=1,\ldots,n$, where $\delta_{ii}=1$ and $\delta_{ij}=0$ for $i\neq j$.

Example Let $e_1,\ldots,e_n$ be the canonical basis of $F^n$, then the dual basis of $(F^n)'$ is $f_1,\ldots,f_n$, where $f_i(a_1,\ldots,a_n)=a_i$.

My question is: Is there any other $f_i(a_1,\ldots,a_n)$ such that $f_i(e_j)=\delta_{ij}$?

Note: $a_i\in F$.

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2 Answers 2

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You have $$f_j(a_1,\ldots,a_n) = f_j(\sum_{i=1}^na_ie_i) = \sum_{i=1}^na_if_j(e_i) = \sum_{i=1}^n a_i\delta_{ji} = a_j.$$

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  • $\begingroup$ I agree with you, actually I asked the wrong question :-), sorry for that. See the edited version. $\endgroup$
    – utobi
    May 7, 2021 at 10:36
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Since $f_i(a_1,\ldots,a_n)=a_i$, you have:

  • $f(e_i)=1$, since $e_i=a_1e_1+\cdots+a_ne_n$, eith $a_j=0$ if $j\ne i$ and $a_i=1$;
  • $f(e_j)=0$ if $j\ne i$, for the same reason

and therefore $f_i(e_j)=\delta_{ij}$.

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  • $\begingroup$ I agree, actually I asked the wrong question :-) sorry for that. See the edited version. $\endgroup$
    – utobi
    May 7, 2021 at 10:35

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