A slight variation on what micromass said:
If you have the d.e. y'= Ay, and B is such that [itex]B^{-1}AB= J[/itex], the "Jordan form" for A, then we can multiply both sides of the equation by [itex]B^{-1}[/itex]:
[tex](B^{-1}y)'= B^{-1}Ay= B^{-1}A(BB^{-1})y= (B^{-1}AB)(B^{-1}y)[/tex]
and, letting [itex]x= B^{-1}y[/itex] write the equation as
[tex]x'= Jx[/itex].<br />
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Now, since J is a Jordan form matrix, that gives a system of "almost uncoupled" equations. If we write x as a column matrix<br />
[tex]\begin{bmatrix}x_1 \\ x_2 \\ ---\\ x_n\end{bmatrix}[/tex]<br />
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Then every equation is of the form [itex]x_k'= \lambda_kx_k[/itex] or [itex]x_k'= \lambda_nx_k+ x_{k+1}[/itex]. The very last equation is, of course, [itex]x_n'= \lambda_n x_n[itex], because there is no "[itex]x_{n+1}[/itex]" and is easy to solve. Then, if [itex]x_{n-1}'= \lambda_{n-1}x_{n-1}+ x_n[/itex], because we already have [itex]x_n[/itex], that is easy to solve, etc.<br />
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Finally, since [itex]x= B^{-1}y[/itex], [itex]y= Bx[/itex].<br />
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If a matrix happens to be "diagonalizable" (all those "1" above the diagonal are unecessary), we can completely "uncouple" all those equations, whether differential equations or not, and have, basically, n separate, simpler, problems. If the matrix is NOT diagonalizable, we can still simplify as much as possible using the "Jordan Normal Form".[/itex][/itex][/tex]