In Linear Algebra Done Right by Axler, there are two sentences he uses to describe the uniqueness of Linear Maps (3.5) which I cannot reconcile. Namely, whether the uniqueness of Linear Maps is determined by the choice of 1) basis or 2) subspace. These two seem like very different statements to me given there can be a many-to-one relationship between basis and subspace. In otherwords, saying a Linear Map is "unique on a subspace" seems like a stronger statement than saying it is "unique on a basis".
This first sentence he writes before proving the theorem (3.5):
The uniqueness part of the next result means that a linear map is completely determined by its
values on a basis.
This second sentence he writes at the end after proving the uniqueness of a linear map:
Thus $T$ is uniquely determined on $span(v_1, \dots, v_n)$ by the equation above.
Because $v_1, \dots, v_n$ is a basis of $V$, this implies that $T$ is uniquely determined
on $V$.
My question is, if $T$ is uniquely determined on $V$, doesn't that imply that the choice of basis for $V$ doesn't matter (since each basis of $V$ spans $V$)? But if so, a key part of the theorem requires explicitly choosing $T$ such that $T(v_j)=w_j$, meaning if we choose a different basis $a_1, \dots, a_n$ of $V$ and then select $T$ such that $T(a_j)=w_j$, we would get a different $T$.
I've found these questions here and here that are tangentially related but don't address my question specifically. On the other hand the questions here and here get a little closer, but the answer in the first suggests that the choice of basis is arbitrary where as the answer in the second suggests the basis must be the same.
For more information, I've included the complete statement of the theorem plus the last paragraph of the proof.
Theorem 3.5 Linear maps and basis of domain
Suppose $v_1, \dots, v_n$ is a basis of $V$ and $w_1, \dots, w_n \in W$. Then there exists a unique linear map $T: V \to W$ such that
$Tv_j=w_j$ for each $j = 1, \dots, n$.
Last paragraph of proof:
To prove uniqueness, now suppose that $T \in \mathcal{L}(V,W)$; and that $Tv_j=w_j$ for each $j = 1, \dots ,n$. Let $c_1, \dots, c_n \in F$. The homogeneity of $T$ implies that $T(c_jv_j) = c_jw_j$ for each $j=1, \dots, n$. The additivity of $T$ now implies that $T(c_1v_1 + \cdots + c_nv_n) = c_1w_1 + \cdots + c_nw_n$. Thus $T$ is uniquely determined on $span(v_1, \dots, v_n)$ by the equation above. Because $v_1, \dots, v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$.
Best Answer
Not an answer, just too long for a comment.
Suppose I have a transformation $T: \mathbb{R}^2 \to \mathbb{R}$.
Pick a basis $B=\{ e_1=(1,0), e_2=(0,1) \}$.
Then $T$ is completely specified by the values $w_1=T(e_1),w_2=T(e_2)$.
Now suppose I pick another basis $C=\{ c_1=(1,1), c_2=(1,-1) \}$.
If I pick values $v_1,v_2$ corresponding to the points $c_1,c_2$ then in order that they match $T$, we must have $v_1 = T(e_1)+T(e_2)$ and $v_2=T(e_1)-T(e_2)$ otherwise we will be specifying a different transformation.
In general you can't just change the basis and keep the same 'w's.