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Let $V$ be a finite-dimentional vector space over the field $F$, and let $B=\{ \alpha_1 ,\alpha_2,....,\alpha_n\}$ be a basis for $V$. Then there is a unique dual basis $B^*=\{ f_1,f_2,....,f_n\} $ for $V^*$ such that $f_i(\alpha_j)=\delta_{ij}.$

My question is why there is a unique basis which is dual to $B$.(why unique?) I know that $f_i(\alpha_j)=\delta_{ij}$ is a basis for dual space but there can be other basis for the dual space!

Thanks!

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Well, if you already found one then observe each $\;f_i\in V^*\;$ is uniquely determined by its values on the basis $\;B\;$ , and this unique determination corresponds with the basis elements indexwise.

Thus, any other such basis must contain linear functionals $\;g_i\in\ V^*\;$ each of which agree with a correspondent $\;f_i\;$ on the basis $\;B\;$ indexwise and thus it is the same dual basis as then $\;g_i=f_i\;$ for all $\;i\;$ .

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  • $\begingroup$ I don't get this point "Thus, any other such basis must contain linear functionals $g_i∈ V^*$ each of which agree with a correspondent $f_i$ on the basis $B$ indexwise and thus it is the same dual basis as then $g_i=f_i$ for all $i$ ". $\endgroup$ – uuuuuuuuuu Mar 29 '16 at 19:51
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    $\begingroup$ @Saikat A dual basis for $\;B\;$ is a basis $\;\{\phi_i\}\;$ of $\;V^*\;$ which fulfills $\;\phi_k\alpha_m=\delta_{km}\;$ , which is a very precise relation: $\;\phi_1\;$ sends precisely $\;\alpha_1\;$ to $\;1\;$ and all the others to zero, $\;\phi_2\;$ sends exactly $\;\alpha_2\;$ to $\;1\;$ and all the rest to zero, etc. Any dual basis has to fulfill this. $\endgroup$ – DonAntonio Mar 29 '16 at 19:59
  • $\begingroup$ @Joanperno I am not getting why any dual space has to fulfill this? $\endgroup$ – uuuuuuuuuu Mar 29 '16 at 20:06
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    $\begingroup$ @Saikat Because that exactly the definition of "dual basis"! $\endgroup$ – DonAntonio Mar 29 '16 at 20:06
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    $\begingroup$ @Saikat For finite dimension we have $\;V\cong V^*\;$ , so in the dual space there are as many different basis as in $\;V\;$ , but I think we're confusing things here: THE dual basis in $\;V^*\;$ is with respect to some specific, given basis in $\;V\;$ ...! That's the reason that THE dual basis in $\;V^*\;$ for a specific, given basis in $\;V\;$ is unique $\endgroup$ – DonAntonio Mar 29 '16 at 20:18
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For every basis $B$ for your space you get $B^*$ , a basis for the dual space. If you take a basis of the dual space $B^*=\{ f_1,...,f_2 \}$ so the new basis $$B^{**}=\{ f_1+f_2,f_2,...,f_{n-1},f_n \}$$ This basis doesn't came from a normal basis because the dimension of the image $f_1+f_2$ is 2, so you can't find a basis that $f_1+f_2$ is zero on all of them but one. So $B^{**}$ is not a basis that is created by the "typical" way.

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  • $\begingroup$ All the linear functionals $\;f_i\;$ are either the functional zero or their image is the whole field and thus have dimension 1. This is exactly the same for the functional $\;f_1+f_2\;$ and thus it cannot have image of dimension $\;>1\;$ . $\endgroup$ – DonAntonio Mar 29 '16 at 20:01
  • $\begingroup$ This is wrong. Every basis of $V^*$ is dual to some basis of $V$, in this case $\{\alpha_1, \alpha_2-\alpha_1, \alpha_3, \dots, \alpha_n \}$ is the basis of $V$ corresponding to $B^{**}$. $\endgroup$ – Christoph Jul 16 '20 at 14:13
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Because a dual basis is the basis of dual space, it spans the dual space. We can suppose $B=\{v_1,...,v_n\}$ is a basis of the dual space, and because it is the basis of the dual space, $B$ spans $V'$. And the dual space $V'$ denotes the set of all linear functionals that map some elements in the space $V$ to a scalar. Therefore, $B$ can represent any linear functional in $V'$. For an arbitrary linear functional $T$ in the dual space $V'$, $T$ can be represented by the basis of dual spaces and it must be unique because every different linear combination of linearly independent list is unique, and without loss of generality we can also say that any linear functionals represented by a linear combination of the basis (that are linearly independent) of $V'$ are also unique.

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  • $\begingroup$ This does not answer the (4 year old) question. $\endgroup$ – Christoph Jul 16 '20 at 14:15

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