Linear Algebra with Differential Equations/Heterogeneous Linear Differential Equations/Diagonalization

First of all (and kind of obvious suggested by the title), $\mathbf {A}$ must be diagonalizable. Second, the eigenvalues and eigenvectors of $\mathbf {A}$ are found, and form the matrix $\mathbf {T}$ which is an augemented matrix of eigenvectors, and $\mathbf {D}$ which is a matrix consisting of the corresponding eigenvalues on the main diagonal in the same column as their corresponding eigenvectors. Then with our central problem:

$\mathbf {X} '=\mathbf {AX} +\mathbf {G} (t)$

We substitute:

$\mathbf {TY} '=\mathbf {ATY} +\mathbf {G} (t)$

Then left multiply by $\mathbf {T} ^{-1}$

$\mathbf {Y} '=\mathbf {T} ^{-1}\mathbf {ATY} +\mathbf {T} ^{-1}\mathbf {G} (t)$

As a consequence of Linear Algebra we take the following identity:

$\mathbf {D} =\mathbf {T} ^{-1}\mathbf {AT}$

Thus:

$\mathbf {Y} '=\mathbf {DY} +\mathbf {T} ^{-1}\mathbf {G} (t)$

And because of the nature of the diagonal the problem is a series of one-dimensional normal differential equations which can be solved for $\mathbf {Y}$ and used to find out $\mathbf {X}$.