Repeated Eigenvalues
Differential Equations · Section 7.7 · generated practice set
One eigenvalue, one eigenvector: generalised eigenvectors and the improper node.
Practice this set → Fresh numbers on every attempt. No account needed.
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)
$A - lambda I eq 0$, 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 $$mathbf{x} = c_1mathbf{v}e^{lambda t} + c_2left(mathbf{v}t + mathbf{w} ight)e^{lambda t}.$$
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 $mathbf{x}' = egin{pmatrix} 3 & -1 \ 1 & 1end{pmatrix}mathbf{x}$.
- $ ext{tr}A = 4$, $det A = 3 + 1 = 4$, so $lambda^2 - 4lambda + 4 = (lambda - 2)^2 = 0$ and $lambda = 2$ repeated.
- $A - 2I = egin{pmatrix} 1 & -1 \ 1 & -1end{pmatrix} eq 0$, so the matrix is defective: rank 1, one eigenvector.
- $(A - 2I)mathbf{v} = mathbf{0}$ gives $v_1 = v_2$, so $mathbf{v} = egin{pmatrix} 1 \ 1end{pmatrix}$.
- Generalised eigenvector: $(A - 2I)mathbf{w} = mathbf{v}$ means $w_1 - w_2 = 1$, so take $mathbf{w} = egin{pmatrix} 1 \ 0end{pmatrix}$.
- $mathbf{x} = c_1egin{pmatrix} 1 \ 1end{pmatrix}e^{2t} + c_2left[egin{pmatrix} 1 \ 1end{pmatrix}t + egin{pmatrix} 1 \ 0end{pmatrix} ight]e^{2t}$ — 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. $te^{lambda t} o 0$ 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.