Are the entries in a matrix the coefficients of the associated linear transform? My question is in regards to the matrices and what they represent in terms of linear transformations, specifically are the entries of the matrix the  coefficients of the associated linear equation? 
I've been trying to work backwards from the identity matrix to the underlying linear equation in order to get a better understanding.
Given the standard basis of
$$
w_1 = [1,0] \\
w_2 = [0,1]
$$
And the 2D identity matrix
$$
    \begin{bmatrix}
    1 & 0 \\
    0 & 1 \\
    \end{bmatrix}
$$
Is a linear equation in terms of basis vectors?
$$x = 1 \cdot w_1 + 0 \cdot w_2$$
$$y = 0 \cdot w_1 + 1 \cdot w_2$$
Or just the terms of the function and the basis is implicit?
$$ f(x,y) = 1 \cdot x + 0 \cdot y $$
$$ f(x,y) = 0 \cdot x + 1 \cdot y $$
 A: Yes the entries of the matrix represent the coefficients related to the linear transformation of the basis vectors. Notably the $i^{th}$ column represents the transformed of the $i^{th}$ basis vector.
That’s precisely the way we construct the matrix associated to a given linear transformation and by which we can also prove its unicity.
A: Writing vectors as columns would be more convenient.
$T$ is a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$; $e_1, e_2,...,e_n$ is the basis of $\mathbb{R}^n$; $b_1,b_2,...,b_m$ is the basis of $\mathbb{R}^m$; $T(x)$ is the coordinates of the image of $x \in \mathbb{R}^n$ under the basis $b_1,...,b_m$.
Then, the matrix of $T$ is
$$
M(T) = \begin{bmatrix} T(e_1)&T(e_2)& ... &T(e_n) \end{bmatrix}.
$$
For example, $T$ is the rotation on $\mathbb{R}^2$. We take the basis $\{\begin{bmatrix} 1\\ 0\end{bmatrix},\begin{bmatrix} 0\\ 1\end{bmatrix}\}$.
$$
T(\begin{bmatrix} 1\\ 0\end{bmatrix}) = \begin{bmatrix} cos\theta\\ sin\theta\end{bmatrix},
$$
$$
T(\begin{bmatrix} 0\\ 1\end{bmatrix}) = \begin{bmatrix} -sin\theta\\ cos\theta\end{bmatrix}.
$$
So the matrix is
$$
\begin{bmatrix}
cos\theta & -sin\theta \\
sin\theta & cos\theta
\end{bmatrix}.
$$
