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I just started learning about linear maps and I had a quick question about the following step from a proof outlining the existence of a unique linear map $T: V \rightarrow W$.

Note: $L(V, W)$ below denotes the set of all linear maps from $V$ to $W$.

Suppose $T \in L(V,W)$ and $Tv_j = w_j$ for $j = 1, \dots, n$. Let $c_1, \dots, c_n \in F$. The homogeneity of $T$ implies that $T(c_j v_j) = c_j w_j$ for $j = 1, \dots, n$. The additivity of $T$ now implies that $$T(c_1 v_1 + \dots + c_n v_n) = c_1 w_1 + \dots + c_n w_n.$$

Thus $T$ is uniquely determined on span$(v_1, \dots, v_n)$ by the equation above.

What does it mean above that "$T$ is uniquely determined on span"? I understand that homogeneity and additivity hold for linear maps (by definition) but I have no idea how we followed those properties in the proof with "unique determination of $T$".

Noah Stebbins
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1 Answers1

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It means that if $Q$ is another linear map from $V$ into $W$ and if the condition$$(\forall j\in\{1,2,\ldots,n\}):Qv_j=w_j$$also holds, then $Q=T$.

José Carlos Santos
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  • Gotcha, thanks for the definition @josé. In the proof in question, how did we go from outlining homogeneity and additivity in $T$ to concluding that $T$ is uniquely determined on span? – Noah Stebbins Jan 02 '19 at 16:25
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    If $v\in\operatorname{span}\bigl({v_1,\ldots,v_n}\bigr)$, then $v$ can be written as $v=c_1v_1+\cdots+c_nv_n$, then\begin{align}T(v)&=T(c_1v_1+\cdots+c_nv_n)\&=c_1T(v_1)+\cdots+c_nT(v_n)\&=c_1Q(v_1)+\cdots+c_nQ(v_n)\&=Q(c_1v_1+\cdots+c_nv_n)\&=Q(v).\end{align} – José Carlos Santos Jan 02 '19 at 17:31