It is something a little different. Let [tex]v_1[/tex] denote a vector represented in basis 1. Then to represent this same vector in terms of a different basis, basis 2, we need to find a matrix [tex]T_{1:2}[/tex] that maps any vector representation from basis 1 to basis 2. Thus if we let [tex]v_2[/tex] denote that vector represented in basis 2, then
[tex]v_2 = T_{1:2} \, v_1[/tex].
This means that
[tex]v_1 = T_{1:2}^{-1} \, v_2[/tex]
so the matrix that maps a vector representation from basis 2 to basis one is
[tex]T_{2:1}=T_{1:2}^{-1}[/tex].
Now, if we have a matrix representation of an operator in basis 1, say [tex]A_1[/tex], then it takes a vector represented in basis 1 and maps it to a different vector represented in basis 1. For our example let
[tex]y_1 = A_1 v_1[/tex].
So if we want to represent y in basis 2 we have,
[tex]y_2 = T_{1:2} y_1 = T_{1:2} A_1 v_1 = T_{1:2} A_1 T_{2:1} v_2[/tex].
Hence, if we want to represent the operator in basis 2, the matrix representation must be
[tex]A_2 = T_{1:2} A_1 T_{2:1} = T_{1:2} A_1 T^{-1}_{1:2}[/tex],
and we have
[tex]y_2 = A_2 v_2[/tex]
as required. If you think about what is happening, it should be easy to remember.
Note that most linear algebra books will cover this.
jason