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

First of all (and kind of obvious suggested by the title), ${\displaystyle \mathbf {A} }$ must be diagonalizable. Second, the eigenvalues and eigenvectors of ${\displaystyle \mathbf {A} }$ are found, and form the matrix ${\displaystyle \mathbf {T} }$ which is an augemented matrix of eigenvectors, and ${\displaystyle \mathbf {D} }$ 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:

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

We substitute:

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

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

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

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

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

Thus:

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

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