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Let V be a finite-dimensional vector space over the field F and let $\{\alpha_1,\ldots,\alpha_n\}$ be an ordered basis for V. Let W be a vector space over the same field F and let $\beta_1,\ldots,\beta_n$ be any vectors in W. Then there is precisely one linear transformation T from V into W such that $$\text{T}\alpha_j=\beta_j,\quad j=1,\ldots,n$$

I think the theorem is perfectly fine without the notion of an ordered basis. Is there any specific reason?

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    $\begingroup$ The fact that the basis is ordered is what gives the uniqueness of $T$. $\endgroup$
    – Sammy Black
    Nov 7, 2015 at 7:49
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    $\begingroup$ Not all that long ago, books did not mention ordered bases, though there was often an implicit order, sequences are clearer than sets. Here we could talk about any function from the basis $B$ to $W$ being uniquely extendable. Order is not necessary. $\endgroup$
    – André Nicolas
    Nov 7, 2015 at 7:58

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We can easily state a version of the theorem without mentioning an ordered basis. Let the set $B$ be a basis for $V$, and let $f$ be any function from $B$ to $W$.Then there is a unique linear transformation $T$ from $V$ into $W$ such that $T(b)=f(b)$ for any $b$ in $B$.

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