Linear system and eigenvalues

We look for a solution of the form \(\mathbf x=e^{\lambda t}\mathbf v\): differentiating gives \(\lambda e^{\lambda t}\mathbf v\), and multiplying by \(A\) gives \(e^{\lambda t}A\mathbf v\). They agree exactly when \(A\mathbf v=\lambda\mathbf v\). Each eigenvalue–eigenvector pair gives a solution, and linear combinations of them are still solutions.

If \(A\) has \(n\) distinct eigenvalues, the \(n\) eigenvectors are independent and the general solution is \(\mathbf x=c_1e^{\lambda_1t}\mathbf v_1+\dots+c_ne^{\lambda_nt}\mathbf v_n\). The constants \(c_i\) are fixed by the initial condition, solving \(c_1\mathbf v_1+\dots+c_n\mathbf v_n=\mathbf x(0)\). Guiding example: for \(A=\begin{pmatrix}1&2\\2&1\end{pmatrix}\), \(\lambda=3\) with \((1,1)\) and \(\lambda=-1\) with \((1,-1)\).

The sign of the real part of each \(\lambda\) decides the fate: if all are negative, solutions tend to the origin; if any is positive, some solutions grow. Complex eigenvalues \(\alpha\pm i\beta\) produce oscillations \(e^{\alpha t}\cos\beta t\) and \(e^{\alpha t}\sin\beta t\). A zero eigenvalue gives a constant solution.

Some cases are not covered by this direct method: repeated eigenvalues without enough eigenvectors, which require generalized eigenvectors. A typical mistake is to pair an eigenvalue with the eigenvector of another. This concept prepares the phase plane, where behavior is classified by the trace and determinant of \(A\).