10 1. Autonomous Linear Differential and Difference Equations Sometimes, we call the elements of the real generalized eigenspaces also generalized eigenvectors. For a real eigenvalue μ its algebraic multiplicity co- incides with the dimension nk = dim Ek of the corresponding real generalized eigenspace, for a complex-conjugate eigenvalue pair μ, ¯ the dimension of the corresponding real generalized eigenspace is given by nk = dim Ek = 2m where m is the algebraic multiplicity of μ. It follows that r k=1 Ek = Rd, i.e., every matrix has a set of generalized eigenvectors that form a basis of Rd. 1.3. Solution Formulas The real Jordan form derived in the previous section will allow us to write down explicit formulas for the solution of the differential equation ˙ = Ax. This is due to the fact that for matrices in real Jordan normal form it is possible to compute the matrix exponential. The price to pay is that first the matrix A has to be transformed into real Jordan normal form. Thus we begin by noting what happens to the solutions under linear transformations. Proposition 1.3.1. Let A, B gl(d, R) with B = S−1AS for some S Gl(d, R). Then the solutions y(t, y0) of ˙ = By and x(t, x0) of ˙ = Ax are related by y(t, y0) = S−1x(t, Sy0),t R. Proof. One computes for t R, y(t, y0) = eBty0 = eS−1ASty 0 = S−1eAtSy0 = S−1x(t, Sy0). A consequence of this proposition is that for the computation of expo- nentials of matrices it is sufficient to know the exponentials of Jordan form matrices and the transformation matrix. The following results for the solu- tions of ˙ = Ax are obtained with the notation above from the properties of the real Jordan normal form. Proposition 1.3.2. Let A gl(d, R) with real generalized eigenspaces Ek = Ek(μk) with dimensions nk = dim Ek,k = 1,...,r. (i) Let v1,...,vd be a basis of Rd, e.g., consisting of generalized real eigenvectors of A. If x0 = d i=1 αivi, then x(t, x0) = d i=1 αix(t, vi) for all t R. (ii) Each generalized real eigenspace Ek is invariant for the linear dif- ferential equation ˙ = Ax, i.e., x(t, x0) Ek for all t R, if x0 Ek. If A is a diagonal matrix, the solutions are easily obtained. Example 1.3.3. Let D = diag(μ1,...,μd) be a diagonal matrix. Then the standard basis e1,..., ed of Rd consists of eigenvectors of D and the solution
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