What is meant by the linear mapping $T$ is entirely and uniquely determined on $V$?

Question

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With reference to a similar post here, I cannot fully understand the terminology "$T$ determined on $\V$". To elaborate, the theorem states:

Let $\V$ and $\W$ be vector spaces over $\F$.

Suppose $\v_1, ..., \v_n$ is a basis of $\V$ and $\w_1, ..., \w_n$ be a set of arbitrary vectors in $\W$. Then there exists an unique linear map $T: \V \to \W$ such that

$$ T(\v_j) = \w_j , \quad \forall j = 1, ..., n $$

---

The author mentioned:

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.

I tried to reconcile this with an example from the link above.

Suppose I have a transformation $T: \mathbb{R}^2 \to \mathbb{R}$. At this point I do not know the exact mapping $T$ yet.

Pick a basis $B=\{\e_1=(1,0), \e_2=(0,1) \}$.

Then $T$ is completely specified by the values $\w_1=T(\e_1),\w_2=T(\e_2)$. At this point, I know such a $T$ must exist and be uniquely determined by the theorem above. This is still vague to me, as I do not understand if $T$ is determined by the basis of $\V$ alone (much confusion with the wording of "determined on $\V$".

So I read on and see the elaboration from the post, the OP mentioned that now he understood that once $T$ is determined by $\w_j$ on $\v_j$, then this $T$ is unique on $\V$. My understanding of determined is that we need to know both $\v_1, \v_2$ (the basis vectors of $\V = \mathbb{R}^2$) and the output $\w_1 = T(\e_1), \w_2 = T(\e_2)$. Consequently, we need to pick any vectors $\w_1, \w_2 \in \W$, say $\w_1 = (1, 1), \w_2 = (2, 0)$ then only can we say that this particular mapping $T$ is uniquely and entirely determined by $\w_j$ on $\v_j$?

Is my understanding correct? It will be great if I can have a concrete example to illustrate the two points the author mentioned.

Answer 1

$S, T\in \mathcal{L}(V, W) $

Then $S=T $ iff $Sv=Tv$ ,$\forall v\in V$

Now, $V, W$ are not just ordinary set and $S, T$ not only ordinary mapping. $V, W$ has a rich algebraic structure - vector space and $S, T$ are linear map. This makes our life easy.

$S=T$ iff $ Sv=Tv, \forall v\in \mathcal{B}_V$

Where $\mathcal{B}_V$ is a basis of $V$.

In other words, if two linear map agrees on a basis of the domain space , then the agrees on the whole domain.

Proof is not difficult,

Take, $v\in V $

$v=\sum_{j\in J }c_j v_j$ where $c_j=0 $ for all but finitely many $j$'s and $v_j \in \mathcal{B}_V$

Now, $\begin{align}Tv = T(\sum_{j\in J }c_j v_j) &=\sum_{j\in J }c_jTv_j \\&=\sum_{j\in J }c_jSv_j= Sv \end{align}$

So, if you can specify a map $T$ linearly on $\mathcal{B}_V$, it can be extended uniquely and linearly.

Suppose, $\mathcal{B}_V=\{v_j : j\in J\}$

$T:\mathcal{B}_V \to W $

$Tv_j=w_j$

Then, $\exists \hat{T} \in \mathcal{L}(V,W) $ unique and $\hat{T}|_{\mathcal{B}_V}=T$

$\hat{T}v=\hat{T}\sum_{j=1}^{n} c_jv_j=\sum_{j=1}^{n} c_j \hat{T} v_j=\sum_{j=1}^{n} c_j Tv_j=\sum_{j=1}^{n} c_jw_j=w$

Where, $v=\sum_{j=1}^{n} c_jv_j$ and $w=\sum_{j=1}^{n} c_jw_j$