Eigenvalues and Eigenvectors
Differential Equations · Section 7.3 · generated practice set
Trace and determinant to the characteristic polynomial, and what the sign pattern says about the phase portrait.
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Try one now
Find the eigenvalues of $$A = \begin{pmatrix} -4 & 0 \\ -2 & -3 \end{pmatrix}.$$
Solution
- Characteristic equation: $\det(A - \lambda I) = \lambda^2 - (\text{tr}A)\lambda + \det A = 0$.
- $\text{tr}A = -4 - 3 = -7$ and $\det A = (-4)(-3) - (0)(-2) = 12$.
- So $\lambda^2 + 7\lambda + 12 = 0$.
- Factor: $(\lambda + 4)(\lambda + 3) = 0$, giving $\lambda = -4, -3$.
- For the system $\mathbf{x}' = A\mathbf{x}$, both eigenvalues are negative, so the origin is an asymptotically stable node.
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Method
For a $2\times2$ matrix, go through trace and determinant: $$\lambda^2 - (\text{tr}A)\lambda + \det A = 0.$$ Then solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ for each eigenvector — the two rows are always multiples of one another, so use whichever is simpler.
Definitions and theorems
Worked example
Find the eigenvalues of $A = \begin{pmatrix} 1 & 2 \\ 4 & 3\end{pmatrix}$.
- $\text{tr}A = 4$, $\det A = 3 - 8 = -5$.
- $\lambda^2 - 4\lambda - 5 = 0$.
- $(\lambda-5)(\lambda+1) = 0 \Rightarrow \lambda = -1, 5$.
- Opposite signs, so the origin is a saddle point for $\mathbf{x}' = A\mathbf{x}$.
Common mistakes
- Sign of the trace term. It is $-(\text{tr}A)\lambda$.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.