On the border between nodes and spirals sits the case of one repeated eigenvalue. If the matrix has only one eigenvector direction, it gives only one straight-line solution, and the second needs a factor of time, just as a repeated root needed a factor of x in chapter four. The trajectories then all come in tangent to that single direction, turning part of the way round as they do: a degenerate node. If instead every vector is an eigenvector, the matrix is a multiple of the identity and every solution is a straight ray: a star.

A crowd leaving a stadium through a single gate. People come from every side, but all of them walk the last stretch in the one direction the gate allows.

$x'=-x+y$, $y'=-y$: the repeated eigenvalue $-1$ has the single eigenvector direction $(1,0)$. Every trajectory arrives tangent to the horizontal axis; those from above swing round to come in from the right, and those from below from the left.

If $(\operatorname{tr}A)^{2}=4\det A$ the eigenvalues are equal, $\lambda=\tfrac12\operatorname{tr}A$. If , every vector is an eigenvector and every solution is : a star. Otherwise there is one eigenvector direction , and a second solution uses a vector with ; every solution is the combination below. For the origin is a stable degenerate node and every trajectory arrives tangent to ; for it is unstable.

• $A=\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$: , and , since .

Every $2\times2$ matrix is a point here, its trace across and its determinant up. Below the axis are the saddles; above the parabola $\det A=\tfrac14(\operatorname{tr}A)^{2}$, where the eigenvalues repeat, the spirals; between the two, the nodes; and on the upper half of the vertical axis, the centres. The left half is stable and the right half unstable.

Differentiate by the product rule: the exponential brings down lambda, and the factor t brings down one more copy of v. Apply the matrix instead, using that it sends v to lambda times v and w to lambda times w plus v. Both sides come out as the same three terms, so the function is a solution; its value at time zero, w, is independent of v, so with the straight-line solution it forms a fundamental pair.

Differentiate by the product rule.

Apply the matrix; the exponential and t are numbers.

Use the two equations for v and w.

The two sides are the same three terms.

A repeated eigenvalue with only one eigenvector direction gives one straight-line solution. The second needs a factor of t, as a repeated root did in chapter four.

Great work! You now know how to:

• find the second solution for a repeated eigenvalue
• prove that it solves the system
• tell a degenerate node from a star
• place every kind of equilibrium in the trace–determinant plane