Repeated Eigenvalues
Differential Equations · Section 7.7 · generated practice set
One eigenvalue, one eigenvector: generalised eigenvectors and the improper node.
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Try one now
The matrix $$A = \begin{pmatrix} -1 & -4 \\ 1 & -5 \end{pmatrix}$$ has a repeated eigenvalue. Find it and classify the origin for $\mathbf{x}' = A\mathbf{x}$.
Solution
- $\text{tr}A = -6$ and $\det A = 9$, so $\lambda^2 + 6\lambda + 9 = 0$.
- The discriminant is $0$ — zero, so $\lambda = \dfrac{\text{tr}A}{2} = -3$, repeated.
- $A + 3I = \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix}$ is not the zero matrix, so it has rank 1 and only one independent eigenvector. The matrix is defective.
- The second solution needs a generalised eigenvector: $\mathbf{x} = c_1\mathbf{v}e^{-3t} + c_2\left(\mathbf{v}t + \mathbf{w}\right)e^{-3t}$, where $(A + 3I)\mathbf{w} = \mathbf{v}$.
- With $\lambda < 0$ the exponential beats the factor of $t$, so every trajectory still reaches the origin — an improper node, asymptotically stable. All trajectories enter or leave tangent to the single eigendirection.
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Method
When $\det(A - \lambda I) = 0$ has a double root, there are two possibilities, and they look nothing alike.
Case 1: $A = \lambda I$
Every vector is an eigenvector. The solution is $\mathbf{x} = e^{\lambda t}\mathbf{x}(0)$ and every trajectory is a straight ray — a proper node (or star point). Easy to spot: the off-diagonal entries are zero and the diagonal entries are equal.
Case 2: defective (the usual one)
$A - \lambda I \neq 0$, so it has rank 1 and there is only one independent eigenvector $\mathbf{v}$. One solution is $\mathbf{v}e^{\lambda t}$; the second needs a generalised eigenvector $\mathbf{w}$ solving $$(A - \lambda I)\mathbf{w} = \mathbf{v}.$$ Then $$\mathbf{x} = c_1\mathbf{v}e^{\lambda t} + c_2\left(\mathbf{v}t + \mathbf{w}\right)e^{\lambda t}.$$
Note the shape: it is $\mathbf{v}te^{\lambda t} + \mathbf{w}e^{\lambda t}$, not $\mathbf{w}te^{\lambda t}$. Guessing $\mathbf{w}e^{\lambda t}$ alone fails, exactly as $e^{rt}$ alone failed for a repeated root in 3.4.
The phase portrait
A defective repeated eigenvalue gives an improper node — asymptotically stable if $\lambda < 0$, unstable if $\lambda > 0$. Every trajectory enters or leaves tangent to the single eigendirection, because the $t$ in $\mathbf{v}te^{\lambda t}$ makes that term dominate.
$\mathbf{w}$ is only determined up to adding a multiple of $\mathbf{v}$; any choice gives a correct general solution.
Definitions and theorems
Worked example
Solve $\mathbf{x}' = \begin{pmatrix} 3 & -1 \\ 1 & 1\end{pmatrix}\mathbf{x}$.
- $\text{tr}A = 4$, $\det A = 3 + 1 = 4$, so $\lambda^2 - 4\lambda + 4 = (\lambda - 2)^2 = 0$ and $\lambda = 2$ repeated.
- $A - 2I = \begin{pmatrix} 1 & -1 \\ 1 & -1\end{pmatrix} \neq 0$, so the matrix is defective: rank 1, one eigenvector.
- $(A - 2I)\mathbf{v} = \mathbf{0}$ gives $v_1 = v_2$, so $\mathbf{v} = \begin{pmatrix} 1 \\ 1\end{pmatrix}$.
- Generalised eigenvector: $(A - 2I)\mathbf{w} = \mathbf{v}$ means $w_1 - w_2 = 1$, so take $\mathbf{w} = \begin{pmatrix} 1 \\ 0\end{pmatrix}$.
- $\mathbf{x} = c_1\begin{pmatrix} 1 \\ 1\end{pmatrix}e^{2t} + c_2\left[\begin{pmatrix} 1 \\ 1\end{pmatrix}t + \begin{pmatrix} 1 \\ 0\end{pmatrix}\right]e^{2t}$ — an unstable improper node, since $\lambda = 2 > 0$.
Common mistakes
- Guessing $\mathbf{w}e^{\lambda t}$ for the second solution. It must be $(\mathbf{v}t + \mathbf{w})e^{\lambda t}$; the $t$ rides on $\mathbf{v}$, not $\mathbf{w}$.
- Solving $(A - \lambda I)\mathbf{w} = \mathbf{0}$ for the generalised eigenvector. The right side is $\mathbf{v}$, not zero — that is the whole distinction.
- Hunting for a second eigenvector. In the defective case there is not one. If you find two independent ones, $A = \lambda I$ and the problem is the easy case.
- Worrying that $\mathbf{w}$ is not unique. It is determined only up to a multiple of $\mathbf{v}$; that freedom is absorbed into $c_1$.
- Assuming the factor of $t$ destabilises a decaying solution. $te^{\lambda t} \to 0$ whenever $\lambda < 0$.
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.