# 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 $\mathbf{X}$.