# Insights about $Tv_j=w_j$, the linear maps and basis of domain.

I'm now self-studying the book Linear Algebra Done Right. In chapter 3, I learned the theorem 3.5, Linear maps and basis of domain: Suppose $v_1,...,v_n$ is a basis of V and $w_1,...,w_n∈W$. Then there exists a unique linear map $T:V→W$ such that $$Tv_j=w_j$$ for each $j=1,...,n$.

"The existence part of the next result means that we can find a linear map that takes on whatever values we wish on the vectors in a basis. The uniqueness part of the next result means that a linear map is completely determined by its values on a basis." -- a paragraph before this theorem.

I'm still wondering how to make use of this theorem. It confused me quite a while, so would you please help me, really really appreciate it.

Suppose that $T:V\to W$ is a linear transformation between vector spaces $V$ and $W$. It is natural to ask what $T$ does to the space $V$. However, this seems like a very daunting question --- how can we completely understand what $T$ does to the uncountably many vectors in $V$? It's not feasible to check $T(v)$ for every $v\in V$, so what do we do?
Well, we find a $\textit{basis}$ for $V$, i.e., a set $\{v_1, \dots, v_n\}$ such that every vector $v\in V$ has a unique representation as $c_1v_1 + \cdots + c_nv_n$. Bases are nice because they contain all the information we need about vectors in $V$. So we don't need to check what happens to every vector in $V$ --- it suffices to check what $T$ does to the basis vectors. Once we know that, since $T$ is linear, we can figure out what $T$ does to any vector $v$ by computing: $$T(v) = T(c_1v_1 + \cdots + c_nv_n) = c_1T(v_1) + \cdots + c_nT(v_n)$$ Since we know what $T(v_1), \dots, T(v_n)$ are, we therefore know what $T(v)$ is.
The fact that a linear transformation $T$ is completely determined by how it acts on a basis $\{v_1,\dots, v_n\}$ means that we have a lot of power in constructing linear transformations. If we want to define a new transformation $S:V\to W$, all we need to do is decide what $S$ does to each $v_i$. Once we do that, everything else falls into place (by the same computation above).