he eigenvalues of the matrix A are computed using the characteristic equation
det(A−λI) = det \begin{bmatrix} −λ & -1 \\ 1 & −λ\end{bmatrix}=λ^{2} +1 = 0
We see that λ^{2}=−1, so that λ=±i, where i is the complex number^{2} i = \sqrt {−1}. To determine the eigenvector associated with λ=i, we solve (A−iI)v =0. Row-reducing the appropriate matrix with complex entries just as we would a matrix with real entries, we observe
\begin{bmatrix} −i & -1&0 \\ 1 & −i&0\end{bmatrix}→ \begin{bmatrix} 1 & -i&0 \\-i & −1&0\end{bmatrix} → \begin{bmatrix} 1 & -i&0 \\0 & 0&0\end{bmatrix}
where the first step was achieved by swapping the two rows, while the last step was achieved by computing the row replacement iR_{1}+R_{2}→R_{2}. It follows that any eigenvector v associated with λ = i must have components v_{1} and v_{2} that satisfy v_{1}= iv_{2}. Choosing v_{2}= 1, we see that an eigenvector v corresponding to λ=i is v =[i 1]^{T}. Similar computations with λ=−i show that a corresponding eigenvector is v= [−i 1]^{T}. While we might suggest at this point that
x(t ) = e^{it} \begin{bmatrix} i \\ 1 \end{bmatrix}
a solution to x´= Ax, such a solution involves the complex number i, and is not a real solution to the system. A plot of the direction field for the system reveals further why no real solutions arise directly from the eigenvectors. In particular, if we examine figure 3.12, the direction field and various trajectories exhibit behavior consistent with the fact that the system has no straight-line solutions due to the fact that it has no real eigenpairs: every trajectory appears to be circular.
In this example, we will suspend our work with eigenvalues and eigenvectors and see whether we can determine a solution to the system more directly. If we examine the two equations given in the system x´= Ax, we observe that we are trying to solve the two equations \acute{x_{1}}=−x_{2} and \acute{x_{2}}= x_{1} simultaneously. In particular, we seek two functions x_{1}(t) and x_{2}(t) such that the derivative of the first is the opposite of the second and the derivative of the second is the first. This is a familiar scenario encountered in calculus and we recognize that x_{1}(t)=cos t and x_{2}(t) = sin t form a pair of such functions. Further consideration reveals that the choices x_{1}(t)=−sin t and x_{2}(t)= cos t also satisfy the system.
Our recent observations show that the vector functions
x_{1}(t ) =\begin{bmatrix} cos t \\ sin t\end{bmatrix} and x_{2}(t ) =\begin{bmatrix} -sin t\\ cos t\end{bmatrix}
each form a real solution to x´= Ax; moreover, it is clear that x_{1}(t) and x_{2}(t) are not scalar multiples of one another, and thus these are two linearly independent solutions to the system. Therefore, theorem 3.3.2 implies that the general solution to the given system is
x(t ) =c_{1} \begin{bmatrix} cos t \\ sin t\end{bmatrix} + c_{2} \begin{bmatrix} -sin t\\ cos t\end{bmatrix}
The presence of the sine and cosine functions in the entries of x will also lead to the circular trajectories we expect from the direction field in figure 3.12.
2: A review of key concepts with complex numbers may be found in appendix B.
