Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 2 / 8B (Differential Equations)

The independence of eigenvectors for distinct eigenvalues ensures that such eigenvectors form a linearly independent set. For completeness, apply ${\prod }_{j\ne k}(M-{\lambda }_{j}I)$ to a proposed relation ${\sum }_i{c}_i{\mathbf{e}}_i=0$. It leaves $c_k{\prod }_{j\ne k}({\lambda }_k-{\lambda }_j){\mathbf{e}}_k=0$, so every $c_k=0$. Thus the three eigenvectors form a basis over the relevant field and $\mathbf{x}={\sum }_i{a}_i(t){\mathbf{e}}_i$. Substitution into the linear system of ordinary differential equations gives ${\sum }_i({\dot{a}}_i-{\lambda }_i{a}_i){\mathbf{e}}_i=0$; linear independence implies ${\dot{a}}_i={\lambda }_i{a}_i$. HenceA real matrix may have a complex conjugate pair of eigenvalues. In that case the calculation is over $\mathbb{C}$ and conjugate coefficients yield real solutions; equivalently use real and imaginary parts of the complex modes. No unstated assumption that all three eigenvalues are real is needed.

The three linearly independent eigenvectors still form a basis, so the same coefficient calculation applies to this diagonalizable matrix matrix. The general solution isA repeated eigenvalue alone does not create a polynomial factor in time; that factor is needed only when a nontrivial Jordan block prevents a full eigenvector basis.

Suppose $\mathbf{v}=\alpha {\mathbf{e}}_1+\beta {\mathbf{e}}_2$. Applying $M-{\lambda }_{2}I$ would give ${\mathbf{e}}_2=\alpha ({\lambda }_1-{\lambda }_2){\mathbf{e}}_1$, contradicting linear independence of the two eigenvectors. Therefore the generalized eigenvector $\mathbf{v}$ lies outside their span and $({\mathbf{e}}_1,{\mathbf{e}}_2,\mathbf{v})$ is a basis. Write $\mathbf{x}=a{\mathbf{e}}_1+b{\mathbf{e}}_2+d\mathbf{v}$. Since $M\mathbf{v}={\lambda }_2\mathbf{v}+{\mathbf{e}}_2$, the linear system of ordinary differential equations givesSolving these equations gives $a=A_1{e}^{{\lambda }_{1}t}$, $d=A_3{e}^{{\lambda }_{2}t}$ and $b=(A_2+A_{3}t)e^{{\lambda }_{2}t}$. ThusThe length-two Jordan chain solution for the Jordan chain $({\mathbf{e}}_2,\mathbf{v})$ produces the $t{e}^{{\lambda }_{2}t}$ mode. The three arbitrary constants span all initial data, so this is the full general solution.