If $V$ and $W$ are vector spaces, $\beta=\{v_1, \ldots , v_n\}$ is a finite a basis for $V$ and $\{w_1, \ldots , w_n\}\subset W$, we know there is an unique linear transformation $T:V\rightarrow W$ such that $T(v_i)=w_i$ for $i=1, 2, \ldots , n$ Is this valid when $V$ is not finite-dimensional?
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3$\begingroup$ You mean, if $\beta = \{ v_\alpha : \alpha \in A\}$ is a basis of $V$ and $\{ w_\alpha : \alpha \in A\}$ is a family of elements of $W$, then there is a unique linear $T\colon V\to W$ with $T(v_\alpha) = w_\alpha$ for all $\alpha \in A$? Yes. $\endgroup$– Daniel FischerMar 22, 2014 at 21:59
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2 Answers
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Yes. Let $\beta = \{v_i:i\in I\}$ be a basis for $V$, where $I$ is some index set. This means, by definition, that each $v\in V$ can be written in the form
$$v = \sum_{i\in I}a_iv_i$$ for some unique collection of scalars $\{a_i:i\in I\}$, only finitely many of which are nonzero (so that the sum above is finite). For each $i\in I$, choose an arbitrary $w_i\in W$. Then we can define a linear map $T:V\to W$ by
$$T\left(\sum_{i\in I}a_iv_i\right) = \sum_{i\in I}a_iw_i$$ which is again well-defined since only finitely many of the scalars $a_i$ are nonzero. This is the unique linear map $V\to W$ that sends $v_i$ to $w_i$ for each $i\in I$.
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The fact that linear maps are uniquely determined by their restriction to a basis of the source vector space is general, and not restricted to the finite dimensional case. Of course if $V$ is infinite dimensional, any basis of $V$ will be infinite, and the family of their images $w_\alpha$ that must be specified to determine $T$ will be infinite as well.
answered Mar 22, 2014 at 22:44