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

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

${𝐗}^{\prime }=𝐀𝐗+𝐆(t)$

We substitute:

${𝐓𝐘}^{\prime }=𝐀𝐓𝐘+𝐆(t)$

Then left multiply by ${𝐓}^{-1}$

${𝐘}^{\prime }={𝐓}^{-1}𝐀𝐓𝐘+{𝐓}^{-1}𝐆(t)$

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

$𝐃={𝐓}^{-1}𝐀𝐓$

Thus:

${𝐘}^{\prime }=𝐃𝐘+{𝐓}^{-1}𝐆(t)$

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