I'm pretty dim when it comes to basis and span, please take that into account :)

Help me prove this theorem:

Let $V$, $W$ be vector spaces, also $T:V \rightarrow W$ is a linear transformation.

If $\beta = \{v_1, v_2, ..., v_n\}$ is a basis for $V$ then $R(T) = span(\{T(\beta)\}) = span (\{T(v_1), T(v_2), ..., T(v_n)\})$


This is a simple consequence of two facts:

1) any vector in $V$ is a linear combination of the vectors $\{v_i\}$ in a basis of $V$ . this means that $\forall u \in V$ we have :$u=u_1v_1+u_2v_2+\cdots + u_nv_n$

2) $T$ is a linear transformation. This means that $\forall u \in V$ we have $T(u)=T(u_1v_1+u_2v_2+\cdots + u_nv_n)=u_1T(v_1)+u_2T(v_2)+\cdots+u_nT(v_n)$

Now, by definition, the range of $T$ is $R(T)=\{w:w\in W,w=T(u)\}$ . This means that $w\in R(T) \iff w= u_1T(v_1)+u_2T(v_2)+\cdots+u_nT(v_n)$, but this means exactly that $R(T)$ is the span of $\{v_1,v_2\cdots v_n\}$.


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