Finding the inverse of linear transformation using matrix Assuming I have a linear transformation represented by a matrix with respect to some random bases, how could I find the inverse of the transformation using the matrix representation?
I know I should find the inverse matrix but from there on, I have no clue what to do.
Thanks in advance!
 A: Let $T$ be an invertible linear transformation from an $n$-dimensional vector space to another, $A$ its $(n \times n)$ matrix, $\mathcal B = \{\alpha_1, \dots, \alpha_n\}$ a basis for the domain, $\mathcal B' = \{\beta_1, \dots, \beta_n\}$ a basis for the co-domain, and $\beta = y_1\beta_1 + \cdots + y_n\beta_n$ a vector in the co-domain.
The inverse of $T, T^{-1},$ is related to $A^{-1}$ by $\left[T^{-1}\beta \right]_{\mathcal B} = A^{-1}[\beta]_{\mathcal B'},$ where $[ \cdot ]_{\mathcal B}$ denotes the coordinate matrix relative to the ordered basis $\mathcal B.$  Then with $A_{ij}^{-1}$ representing the $ij$ entry of $A^{-1},$ we compute
$$\begin{align}
\left[T^{-1}\beta \right]_{\mathcal B} & = A^{-1}[\beta]_{\mathcal B'}\\\\
& = A^{-1}[y_1\beta_1 + \cdots + y_n\beta_n]_{\mathcal B'}\\\\
& = A^{-1}\begin{bmatrix}y_1\\ \vdots\\ y_n\end{bmatrix}\\\\
& = \begin{bmatrix}\sum_{i=1}^nA_{1i}^{-1}y_i\\ \vdots\\ \sum_{i=1}^nA_{ni}^{-1}y_i\end{bmatrix};\\\\
T^{-1}\beta & = \alpha_1 \sum_{i=1}^nA_{1i}^{-1}y_i + \cdots + \alpha_n \sum_{i=1}^nA_{ni}^{-1}y_i.\end{align}$$
If you are satisfied having your inverse in terms of $\mathcal B,$ we are done.
If, however, you want the inverse in terms of the standard ordered basis, express each $\alpha_i$ in terms of the standard basis, and group the terms in the last expression above by the elements of the standard basis.
For an example, see my answer to Find linear transformation given its matrix representation.  The procedure there is the same as that of obtaining $T^{-1}$ from $A^{-1}.$
