So, here's the question that I'm stuck on:
Let $V$ and $W$ be two vector spaces with bases $(v_1,v_2,v_3)$ and $(w_1,w_2,w_3)$. Let $f: V \to W$ be the linear map such that $f(v_i) = w_i$. Find the matrix associated with this map.
So, my approach to this starts off with noticing that $\dim(V) = \dim(W)$. They are isomorphic to each other. In fact, the way that $f$ is defined means that it is isomorphic.
Now, consider the canonical basis isomorphism $h_1: F^3 \to V$. That means that $h_1 (e_i) = v_i$. Similarly, there is a canonical basis isomorphism $h_2: F^3 \to W$. That means that $h_2(e_i) = w_i$.
Then:
$f(v_i) = f(h_1 (e_i)) = h_2(e_i)$
Now, given the fact that $f \circ h_1 = h_2 \circ A$, where A is the associated matrix, it stands to reason that $A$ is associated with the identity transformation and so, $A$ is just the 3 x 3 identity matrix.
Is my reasoning above correct? How can I modify it so it's better? This topic has sort of not been covered in the greatest detail in the book I'm using, with no examples given in the text so I'm not quite sure if my approach makes sense or not.
Edit
So over here, the author clearly knows that $A$ represents a matrix but he seems to put $f$ and $A$ on the same footing, even making it very clear in the commutative diagram that they’re equal.
My idea for the solution above came from the fact that the commutativity of the diagram just means that:
$f \circ h_1 = h_2 \circ A$
In this case, I’ve just explicitly treated $A$ as the linear map between $F^3$ and $F^3$.
"Find the matrix associated with this map". If that's literally the assignment given to you, the instructor should be yelled at for a significant amount of time.
You can only talk about the matrix of a linear map $T\colon V\to W$ with respect to chosen bases $\alpha=\{v_1, \dots ,v_n\}$ of $V$ and $\beta=\{w_1,\dots ,w_m\}$ of $W$. Such a matrix is defined as follows: As $\beta$ is a basis, for each $1\leq j\leq n$, there exists a unique expression of the form $T(v_j)=\sum_{i=1}^ma_{ij}w_i$. The matrix $A\in \mathbb{R}^{m\times n}$ defined by $A_{ij}=a_{ij}$ is the matrix of $T$ with respect to the bases $\alpha$ and $\beta$. I like to denote this matrix by $[T]_{\alpha}^{\beta}$.
Thus in your particular case, $[f]_{\alpha}^{\beta}=\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix}$.
NOTE: I follow the convention that the linear map assigned to a matrix is given by left action. Thus, if $A\in \mathbb{R}^{n\times m}$, it yields a linear map $\mathbb{R}^m\to \mathbb{R}^n\colon X\mapsto AX$ where $X$ is viewed as a column vector.