Why matrix representation of a linear transformation doesn't encode choice of basis for its range and domain? A linear transformation T is defined to be:
$T: V \mapsto W$, $Tv_k= \sum A_{i,j} *w_j$, where $v_i$'s are basises for $v$ and $w_j$'s are basises for $W$.
A matrix representation of $T$ only encodes the $A_{i,j}$ leaving the choice of basis for domain and range( $v_i$'s and $v_j$'s).
To me it sounds like a big flaw. Choice of basis for $V$ and $W$ affects $T$'s characteristics like being Identity of not.
For example if $V$ and $W$ are both $\mathbb{R^2}$,  $v_1 = (1,0)$, $v_2 =(1,0)$ and $w_1 =  (0, 1)$ and $w_2 = (1,0)$, let
\begin{equation}
T = \begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix} .
\end{equation}
$T$ is going to map elements of $V$ to same elements of $W$. Thus, $T$ is identity while it doesn't look like a identity matrix.
Similarly, an Identity matrix where choice of basis for its domain and range are not the same, is not going to be identity.
This was annoying me for a while until I pinpointed the problem.
 A: This ‘problem’ that annoys you is a very common issue in mathematics: to confuse an object with its possible representations.
In this case, a linear transformation $T$ between vector spaces is your object, and different choices of basis for its domain and range yield different representations of it. Of course, nothing prevents that two different linear maps have the same associated matrix in (necessarily different!) representations.
Note that, for example, if your $T$ is an isomorphism, and you take as basis $\{e_1,…,e_n\}$ and $\{T(e_1),…,T(e_n)\}$, then its matrix representation is the identity matrix, even if the domain and range are different spaces (so the linear map cannot be the identity).
This is why you cannot identify a linear transformation with a matrix, but with a whole set of matrices each of which is related to the other matrices in the set by a similarity transformation (that is, by a ‘change of basis’).
The moral of all this is: always keep in mind that the important thing is $\textit{the object}$ you’re working with, while its $\textit{representations}$ are just tools that enable us to work and make computations, but that don’t determine the object by their own.
A: If we have $V = W$, it is very uncommon to use different bases for $V$ and for $W$, and if we are using the same basis for $V$ and for $W$, the identity transformation is always represented by the identity matrix. And if $V\neq W$, there is no identity transformation anyways, so in that case the particular issue that you're highlighting is irrelevant.
If we want to use numbers to do calculations on vector spaces, we more or less need to choose a basis. This is an inherent property (some might even say flaw) in how modern linear algebra works. We can be clever when we make this choice, and we can be stupid. If we find that the choice that was easy in one part of our problem makes another part of our problem difficult, we can change basis.
If you have $V = W$ and are working with linear transformations $T:V\to W$, but for some reason want a different basis for $V$ than for $W$, then yes, that is going to make the identity transformation not look like the identity matrix. This is just a trade-off you are going to have to live with, presumably because your choice to use different bases makes some other part of your problem simpler. If you prefer, you can separate your calculations in one linear transformation $V\to W$ using the same bases, then a change-of-basis matrix from $W$ to get into the basis you are after. It is entirely up to you which of these approaches work best.
