If A is a linear transformation from vector space V to itself and {e1, e2, ..., en} is a basis for V, you can find the matrix representation for A by applying to each of the basis vectors in turn. For example, suppose the matrix representation of A, in that basis, is
\left[\begin{array}{ccc}a_{11} & a_{12} & a{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a{32} & a_{33}\end{array}\right]
then, since the array representation of the basis vectors is just [1, 0, 0], [0, 1, 0], [0, 0 , 1], Applying A to e1 is just
\left[\begin{array}{ccc}a_{11} & a_{12} & a{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a{32} & a_{33}\end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ 0\end{array}\left]= \left[\begin{array}{c} a_{11} \\ a_{21} \\ a_{31}\end{array}\right]
That is, applying A to e1 tells us what the first column is, e2, the second column, etc.
The standard basis for R2 is {[1 0], [0 1]}. Applying A to [1 0] gives [0 -1] and applying A to [0 1] is [0 -1] and to [0 1], [-1 0] so the matrix is
\left[\begin{array}{cc} 0 & -1 \\-1 & 0\end{array}\right]
However, if we use basis {[0 1],[1 0]}, that is the same vectors but in different order, that swaps the columns for A:
\left[\begin{array}{cc} -1 & 0 \\0 & -1\end{array}\right]
In the first example, e1= [1 0], e2= [0 1]. In the second example, e1= [0 1], e2= [1 0].
Same vector space, same linear transformation, same vectors in each basis but, because the order or basis vectors is different, different bases and so different matrices.
