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;Linear differential equations
:The differential equation ''x' '' = ''Ax'', ''x''(0) = ''a'' has the solution ''x''(''t'') = exp(''t A'') ''a''. For a symmetric matrix ''S'', it follows that <math> x(t) = E^T \mbox{Diag} (\exp t e) E a </math>. If <math> a = \sum_{i = 1}^n a_i E_i </math> is the expansion of ''a'' by the eigenvectors of ''S'', then <math> x(t) = \sum_{i = 1}^n a_i \exp(t e_i) E_i </math>.
:Let <math> W^s </math> be the vector space spanned by the eigenvectors of ''S'' which correspond to a negative eigenvalue and <math> W^u </math> analogously for the positive eigenvalues. If <math> a \in W^s </math> then <math> \mbox{lim}_{t \ \infty} x(t) = 0 </math> i.e. the equilibrium point 0 is attractive to ''x''(''t''). If <math> a \in W^u </math> then <math> \mbox{lim}_{t \ \infty} x(t) = \infty </math>, i.e. 0 is repulsive to ''x''(''t''). <math> W^s </math> and <math> W^u </math> are called ''stable'' and ''unstable'' manifolds for ''S''. If ''a'' has components in both manifolds, then one component is attracted and one component is repelled. Hence ''x''(''t'') approaches <math> W^u </math> as <math> t \to \infty </math>.
== Generalizations ==
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