A slight variation on what micromass said:
If you have the d.e. y'= Ay, and B is such that B^{-1}AB= J, the "Jordan form" for A, then we can multiply both sides of the equation by B^{-1}:
(B^{-1}y)'= B^{-1}Ay= B^{-1}A(BB^{-1})y= (B^{-1}AB)(B^{-1}y)
and, letting x= B^{-1}y write the equation as
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 />
\begin{bmatrix}x_1 \\ x_2 \\ ---\\ x_n\end{bmatrix}<br />
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Then every equation is of the form x_k&#039;= \lambda_kx_k or x_k&#039;= \lambda_nx_k+ x_{k+1}. The very last equation is, of course, x_n&#039;= \lambda_n x_n, because there is no &amp;quot;x_{n+1}&amp;quot; and is easy to solve. Then, if x_{n-1}&amp;amp;#039;= \lambda_{n-1}x_{n-1}+ x_n, because we already have x_n, that is easy to solve, etc.&lt;br /&gt;
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Finally, since x= B^{-1}y, y= Bx.&lt;br /&gt;
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If a matrix happens to be &amp;quot;diagonalizable&amp;quot; (all those &amp;quot;1&amp;quot; above the diagonal are unecessary), we can completely &amp;quot;uncouple&amp;quot; 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 &amp;quot;Jordan Normal Form&amp;quot;.