By John W. Cain, Angela M. Reynolds

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**Extra info for Ordinary and Partial Differential Equations: An Introduction to Dynamical Systems **

**Example text**

1 1 0 0 0 0 Hence, solutions of ( A − 2I )v = 0 have the form v1 = v2 = 0, with v3 free. It follows that 0 0 1 is an eigenvector for λ = 2. Unfortunately, since the geometric multiplicity of this eigenvalue is only 1, we have failed to produce a set of 3 linearly independent eigenvectors for the matrix A, which means that A is not diagonalizable. 35 suggests that we compute a generalized eigenvector for λ = 2. We must solve ( A − 2I )2 v = 0, or equivalently 1 0 0 v1 0 1 0 0 v2 = 0 .

Eigenvectors corresponding to different eigenvalues are linearly independent. Proof. We prove this statement for a set of 2 eigenvectors; the reader can extend the proof to the general case. Let v1 and v2 be eigenvectors of a matrix A corresponding to different eigenvalues λ1 and λ2 . Suppose indirectly that these two eigenvectors are linearly dependent. Then there exists a constant c such that v2 = cv1 . Moreover, since eigenvectors are non-zero, it must be the case that c = 0. Multiplying both sides of the equation by A, we have Av2 = cAv1 .

Above, we did not consider the possibility that the matrix M in the equation x = Mx has λ = 0 as an eigenvalue. In such cases, the origin is called a degenerate equilibrium. Notice that if λ = 0 is an eigenvalue, then det M = 0, which means that M is a singular matrix. It follows that the solutions of the equation Mx = 0 form a subspace of dimension at least 1, and any vector x in this subspace would be an equilibrium for our system of ode s. In the remainder of the course, we will typically work only with systems which have isolated equilibrium points (defined later), as opposed to systems with infinitely many equilibrium points.