Question 3.5.4: Let A =[-1 -2;2 -1] and consider the system of differential ...

For matrices with complex eigenvalues, Maple provides an efficient and valuable approach: the program completes the necessary complex arithmetic automatically and produces the results we need. Doing so, we find that A has complex eigenvalues λ = −1 ± 2i with corresponding complex eigenvectors v = [±i   1]^{T}. We choose one of these complex eigenpairs and consider the complex function

z(t ) = e^{(−1+2i)t} \begin{bmatrix} i \\ 1 \end{bmatrix}

Observe that e^{(−1+2i)t} = e^{−t} e^{2ti} , so by Euler’s formula

e^{(−1+2i)t} = e^{−t} (cos2t +i sin2t )

Substituting this fact into z(t ), we observe that

z(t ) = e^{−t} (cos2t +i sin2t ) \begin{bmatrix} i \\ 1 \end{bmatrix}

= e^{−t} \begin{bmatrix}−sin2t +i cos2t \\ cos2t +i sin2t \end{bmatrix}

= e^{−t} \begin{bmatrix}−sin2t  \\ cos2t \end{bmatrix}+ie^{−t}\begin{bmatrix}cos 2t  \\ sin2t \end{bmatrix}

By theorem 3.5.2, it now follows that the real and imaginary parts of z(t ) form two real linearly independent solutions to x´= Ax, and therefore the general solution to x´= Ax is

x(t ) = c_{1}e^{−t} \begin{bmatrix}−sin2t +i cos2t \\ cos2t +i sin2t \end{bmatrix} +c_{2}e^{−t} \begin{bmatrix}cos 2t  \\ sin2t \end{bmatrix}

Since A is an invertible matrix, the origin is the only equilibrium solution of the system. Finally, as figure 3.13 shows, the direction field and plotted trajectories exhibit behavior consistent with the fact that the system has no real eigenvectors and therefore no straight-line solutions. Moreover, since the real part of λ=−1+2i is negative, the role of e^{−t} in the general solution (3.5.16) draws every solution to 0 and thus the origin is a stable equilibrium.