basis of domain — Jemoka Knowledge Base

Suppose v_1, \dots v_{n} \in V is a basis of some vector space V; w_1, \dots w_{n} \in W is just a good’ol list of length n= \dim V in W. There exists a unique linear map T \in \mathcal{L}(V,W) such that…

\begin{equation} Tv_{j} = w_{j} \end{equation}

for each j = 1, \dots n Intuition The layperson’s explanation of this result: 1) that, for everywhere you want to take the basis of one space, there’s always a unique linear map to take you there. 2) that, a linear map is determined uniquely by what it does to the basis of its domain. Proof We have two vector spaces, V and W; v_1, \dots v_{n} \in V forms a basis of V; w_1, \dots w_{n} \in W are just some vectors in W. Definition We define some T: V \to W as follows:

\begin{equation} T(c_1v_1 + \dots + c_{n}v_{n}) = c_1 w_1 + \dots + c_{n} w_{n} \end{equation}

where, c_1, \dots c_{n} \in \mathbb{F}. Note that the actual values of c doesn’t actually matter here. Existence We now show that the T defined above has the property of mapping Tv_{j} \to w_{j}. As the basis v_1, \dots v_{n} is a spanning list of V, some T that takes an arbitrary linear combination of v as input does indeed have domain V. Due to addition’s closure, a linear combination of w is \in W. This makes T at least a function from V \to W. Of course, by taking all c_{i} to 0 except for the index c_{j} you are interested in to 1, you can show that this T takes v_{j} to w_{j}. We now show that T is a Linear Map. This part proof is just route algebra so I won’t type it again. Uniqueness Suppose there is a Linear Map that has the desired property: that T \in \mathcal{L}(V,W) and that Tv_{j}=w_{j}, \forall j=1, \dots n. For any scalar c_{j}, the homogeneity of T indicates that this same T has to take T(c_{j}v_{j}) = c_{j}Tv_{j} = c_{j}w_{j}. Now, the additivity of T also indicates that we can string these c_{j} v_{j} together in the same T; that: given T(c_{j}v_{j}) = c_{j}w_{j}, we can just string it all together to get T(c_1v_1 + \dots + c_{n}v_{n}) = c_1w_1+ \dots + c_{n}w_{n}. This means that there is only one T that behaves in the way that we desire, on the span of v_1 \dots v_{n}. Those vectors being the basis, their span is just the domain V. This makes T uniquely determined on V as we were able to construct the original given map simply by following the rules of the Linear Map.