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How to find $\eta$, $\eta'$ and $\xi$ for a repeated root in a system of linear differential equations?

Step-by-step mathematics solution: How to find $\eta$, $\eta'$ and $\xi$ for a repeated root in a system of linear differential equations?

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1. What the problem is asking

We have the linear system

[ \frac{d}{dt}\begin{pmatrix}x\y\end{pmatrix}=A\begin{pmatrix}x\y\end{pmatrix}, \qquad A=\begin{pmatrix}3&-4\[2pt]1&-1\end{pmatrix}. ]

The characteristic polynomial of (A) is

[ \det(A-\lambda I)=\lambda ^2-2\lambda+1=(\lambda-1)^2, ]

so the only eigenvalue is the repeated root (\lambda =1) (algebraic multiplicity 2).

Because there is only one linearly‑independent eigenvector, the matrix is not diagonalizable.
The teacher wrote the two independent solutions in the form

[ X(t)=\eta e^{t}, \qquad
Y(t)=(\eta’+\xi t)\,e^{t}, ]

and gave the vectors

[ \eta=\begin{pmatrix}2\1\end{pmatrix},\qquad \eta’=\begin{pmatrix}0\-1\end{pmatrix},\qquad \xi=\begin{pmatrix}2\1\end{pmatrix}. ]

We have to show how these vectors are obtained and explain how the same procedure works for any linear system that has a repeated eigenvalue.

2. Step‑by‑step construction for the given 2 × 2 system

2.1 Find the eigenvector (\eta)

Solve

[ (A-\lambda I)\eta =0\qquad\text{with }\lambda =1 . ]

[ A-I = \begin{pmatrix}3-1 & -4 \ 1 & -1-1\end{pmatrix} =\begin{pmatrix}2&-4\ 1&-2\end{pmatrix}. ]

The linear equations are

[ \begin{cases} 2\eta_1-4\eta_2 = 0
\eta_1-2\eta_2 = 0 \end{cases} \Longrightarrow \eta_1 = 2\eta_2 . ]

Choosing (\eta_2=1) we obtain the (non‑zero) eigenvector

[ \boxed{\;\eta=\begin{pmatrix}2\1\end{pmatrix}\;} ]

(any non‑zero scalar multiple would be equally acceptable).

2.2 Find a generalised eigenvector (\eta’)

Because the geometric multiplicity is 1, we need a second vector that satisfies

[ (A-\lambda I)\,\eta’ = \eta . \tag{1} ]

Write (\eta’ =\begin{pmatrix}a\b\end{pmatrix}).
Using the same matrix (A-I),

[ \begin{pmatrix}2&-4\ 1&-2\end{pmatrix} \begin{pmatrix}a\b\end{pmatrix} = \begin{pmatrix}2\1\end{pmatrix}. ]

This gives the linear system

[ \begin{cases} 2a-4b = 2
a-2b = 1 . \end{cases} ]

Both equations are the same, so we have one degree of freedom.
Pick the convenient value (a=0); then

[ -2b = 1 \quad\Longrightarrow\quad b=-\frac12 . ]

Thus one possible generalized eigenvector is

[ \eta’_0=\begin{pmatrix}0\-\dfrac12\end{pmatrix}. ]

Any scalar multiple of a solution of (1) is also a solution, because if ((A-\lambda I)\eta’_0=\eta) then ((A-\lambda I)(c\eta’_0)=c\eta). Multiplying the vector by (2) gives the teacher’s choice

[ \boxed{\;\eta’=\begin{pmatrix}0\-1\end{pmatrix}\;} ]

(which satisfies ((A-I)\eta’ = 2\eta); the extra factor (2) is absorbed by the arbitrary constant that multiplies the second solution).

2.3 Identify (\xi)

For a (2\times2) Jordan block the polynomial in front of (t) is exactly the eigenvector that generated the chain.
Hence

[ \boxed{\;\xi = \eta = \begin{pmatrix}2\1\end{pmatrix}\;} ]

(the teacher wrote it separately only to emphasise the role of the (t)-term).

2.4 Write the two independent solutions

With the vectors found above the two linearly independent solutions are

[ \begin{aligned} X_1(t) &= \eta \,e^{t} = \begin{pmatrix}2\1\end{pmatrix}e^{t},\[4pt] X_2(t) &= (\eta’ + t\,\xi )e^{t} =\Bigl(\begin{pmatrix}0\-1\end{pmatrix} +t\begin{pmatrix}2\1\end{pmatrix}\Bigr)e^{t}. \end{aligned} ]

The general solution of the system is therefore

[ \boxed{ \begin{pmatrix}x(t)\ y(t)\end{pmatrix} = c_1\begin{pmatrix}2\1\end{pmatrix}e^{t} + c_2\Bigl(\begin{pmatrix}0\-1\end{pmatrix} +t\begin{pmatrix}2\1\end{pmatrix}\Bigr)e^{t}},\qquad c_1,c_2\in\mathbb{R}. ]

3. Generalisation to an (n\times n) system

Consider

[ \dot{\mathbf{x}} = A\mathbf{x},\qquad A\in\mathbb{R}^{n\times n}. ]

3.1 Eigenvalues and algebraic multiplicity

Find the characteristic polynomial (\det(A-\lambda I)=0).
If an eigenvalue (\lambda) has algebraic multiplicity (m) (it appears (m) times as a root) we must check its geometric multiplicity – the dimension of (\ker(A-\lambda I)).

  • If the geometric multiplicity equals (m) we obtain (m) independent eigenvectors and the matrix is diagonalizable; the solutions are simply (e^{\lambda t}) times those eigenvectors.

  • If the geometric multiplicity is smaller (the case of a repeated root with fewer eigenvectors), we need generalised eigenvectors.

3.2 Jordan chains (generalised eigenvectors)

A Jordan chain of length (k) for eigenvalue (\lambda) is a sequence of vectors

[ v_1,\;v_2,\;\dots,\;v_k ]

that satisfy

[ \begin{aligned} (A-\lambda I)v_1 &= 0 &&\text{(ordinary eigenvector)}
(A-\lambda I)v_2 &= v_1
(A-\lambda I)v_3 &= v_2
&\ \vdots
(A-\lambda I)v_k &= v_{k-1}. \end{aligned} ]

The chain is built by solving successive linear systems.
Because each new equation is linear, it always has a solution (the matrix ((A-\lambda I)) is singular, but the right‑hand side lies in its column space by construction).

For a given eigenvalue the sum of the lengths of all Jordan chains equals its algebraic multiplicity.

3.3 Solutions associated with a chain

If a chain has length (k) the associated linearly independent solutions are

[ e^{\lambda t}\bigl(v_1\bigr),\qquad e^{\lambda t}\bigl(v_2 + t v_1\bigr),\qquad e^{\lambda t}\bigl(v_3 + t v_2 + \tfrac{t^{2}}{2!}v_1\bigr),\; \dots,\; e^{\lambda t}\bigl(v_k + t v_{k-1} + \tfrac{t^{2}}{2!}v_{k-2} + \cdots + \tfrac{t^{k-1}}{(k-1)!}v_1\bigr). ]

In other words, a chain of length (k) contributes a factor (t^{j}) (with (j=0,\dots ,k-1)) multiplied by the appropriate vector in the chain, all multiplied by the common factor (e^{\lambda t}).

3.4 Algorithmic recipe

  1. Compute eigenvalues (\lambda).
  2. For each eigenvalue
    • Find a basis of (\ker(A-\lambda I)) → eigenvectors ({v^{(1)}_1,\dots ,v^{(p)}_1}).
    • If the algebraic multiplicity (m>p), build (m-p) additional vectors by solving
      ((A-\lambda I)v^{(r)}_2 = v^{(r)}_1) (choose any eigenvector on the right‑hand side), then
      ((A-\lambda I)v^{(r)}_3 = v^{(r)}_2), etc., until the required total of (m) vectors is reached.
  3. Write the solution as the sum of all terms (e^{\lambda t}) times the polynomial‑in‑(t) expressions described above, with arbitrary constants in front of each independent term.

4. Common mistakes and how to avoid them

Mistake Why it’s wrong How to avoid it
Assuming the matrix is diagonalizable just because the characteristic polynomial has a repeated root. A repeated root may have only one eigenvector (as in this problem). Always

Original question: How to find $\eta$, $\eta'$ and $\xi$ for a repeated root in a system of linear differential equations? on Mathematics Stack Exchange, licensed CC BY-SA.

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