I am self-studying linear algebra done right by Axler and got confused about matrix representations of linear transformations. I feel like my question is really elementary, but the more I think about it the more conused I get. Below is the definition of the matrix of a linear map given by Axler on p70 (3rd ed.)
Suppose $T \in \mathcal{L}(V, W)$ [this is the vector space of linear transformations from $V$ to $W$] and $v_1, \ldots, v_n$ is a basis of $V$ and $w_1, \ldots, w_m$ is a basis of $W$. The matrix of $T$ wrt these bases is the $m$-by-$n$ matrix whose entries $A_{j,k}$ are determined by $Tv_k = A_{1,k} w_1 + \ldots +A_{m,k}w_m$
Now let $T: V \rightarrow V$ and let $\mathcal{B} = \{v_1, \ldots, v_n\}$ and $\mathcal{C} = \{u_1, \ldots, u_n\}$ be bases of $V$. Define $Tv_i = u_i$ for $1 \leq i \leq n$. Then, from what I understand, the matrix of $T$ wrt to $\mathcal{B}$ and $\mathcal{C}$ should have 1's on its diagonal and zeroes everywhere else. So it looks the same as the identity map on $V$. I feel like that interpreting the matrix of $T$ this way is incorrect, but I am unsure why.
Another way to view this is (I think), to use a function
$
\phi_\mathcal{A}: V \rightarrow \mathbb{F}^{n}
$ that outputs a list wrt a basis $\mathcal{A}$ of $V$. Here, $\mathbb{F}$ denotes either the set of complex or real numbers. So, if we write $v$ as a linear combination of the basis $\mathcal{B}$, i.e. $v = a_1 v_1 + \ldots + a_n v_n$ for some $a_1, \ldots, a_n \in \mathbb{F}$, then
$$[v]_\mathcal{B} = \phi_{\mathcal{B}}(v) = \begin{bmatrix}
a_1 \\
\vdots \\
a_n
\end{bmatrix}
$$
Then,
$$
T(v) = I [v]_\mathcal{B} = [v]_\mathcal{C}
$$
This is where I feel like I am reasoning incorrectly. Should I interpret the above equation as that $\phi_\mathcal{C}$ and $\phi_\mathcal{B}$ just output the same object, but represent different vectors wrt their bases? Is it correct that I should interpret $I$ as a change of basis rather than the identity map on $V$? If this is the case, is it always the case that given an invertible matrix we can associate it with either a linear transformation where the basis does not change or a change-of-basis?
Best Answer
I feel that your way of thinking using $\phi_{A}$ is a good one. Namely, if we have a linear map $T : V \to W$ and bases $A$ of $V$ and $B$ of $W$, then the matrix representation of $T$ is the linear map $M_{A}^{B}(T) : \mathbb{R}^n \to \mathbb{R}^m$ defined by $M_{A}^{B}(T) = \phi_{B}^{-1}T\phi_{A}$. In words, $M_{A}^{B}(T)$ takes the coordinates of a vector $v \in V$ with respect to the basis $A$ of $V$ and returns the coordinates of the vector $Tv \in W$ with respect to the basis $B$ of $W$. So your definition of $Tv_i = u_i$ translates directly to $M_{A}^{B}(T)e_i = e_i$.