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Suppose I have a finite-dimensional vector space $V$, and suppose that $(v_1,\ldots,v_n)$ is a basis for $V$. If I define $T:V\to W$, which $W$ is a vector space over the same field as $V$, and for some $w_1,\ldots,w_n\in W$ I define: $T(v_1) = w_1,\ldots,T(v_n) = w_n$, does this necessarily say that $T$ is a linear transformation?

How, for example, do I find $T(v_1)+T(v_2)$ in order to prove that that equals to $T(v_1+v_2)$?

Thanks.

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I think you have some confusion...

You say you define $T:V \longrightarrow W$ as $T(v_i)=w_i$. Actually you just defined a map $T: \{ v_1, \dots, v_n\}\longrightarrow W$. So it makes no sense to wonder wether it is linear or not, since $\{ v_1, \dots, v_n\}$ is not a vector space.

However, $\{ v_1, \dots, v_n\}$ is a basis for $V$, so every element $v \in V$ can be written in a unique way in the form $a_1v_1 + \dots +a_nv_n$. So, if you want to extend your definition of $T$ on the whole $V$ such that $T$ is linear, you need to define $$T(a_1v_1 + \dots +a_nv_n) = a_1T(v_1) + \dots + a_nT(v_n) = a_1 w_1 + \cdots + a_nw_n$$

In this way, you have $T:V \longrightarrow W$ linear by construction.

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You will have to specify more conditions to define a transformation. And to prove linear, you can show $T(cv_1+v_2)=cT(v_1)+T(v_2)$.

A counter example is: let $V=W=\mathbb{R}^2$, the basis be the standard basis for both. Define $T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x^2\\y^2\end{pmatrix}$. This transformation maps $\begin{pmatrix}1\\0\end{pmatrix}$ to $\begin{pmatrix}1\\0\end{pmatrix}$ and maps $\begin{pmatrix}0\\1\end{pmatrix}$ to $\begin{pmatrix}0\\1\end{pmatrix}$, but it is not linear.

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  • $\begingroup$ My question is, suppose that $V = Span(v_1, v_2)$. If I define $T(v_1) = w_1, T(v_2)=w_2$ is that enough to define $T$ as a linear transformation? $\endgroup$
    – JonTrav
    Commented Apr 2, 2015 at 9:21
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    $\begingroup$ What I meant is there are many possibilities of $T$ when you only define it by the basis. The example I gave you satisfies your two conditions. There can be many others satisfying your conditions. They could be linear or nonlinear. I think Crostul's answer explained this well. $\endgroup$
    – KittyL
    Commented Apr 2, 2015 at 9:26
  • $\begingroup$ Very nice addition to Crostul's answer; I never would've caught this on my own! $\endgroup$
    – Rax Adaam
    Commented Dec 10, 2020 at 19:32

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