# ODE system, case algebraic multiplicity = 3 , geometric multiplicity = 2

I am solving a ODE system with k = 3 (algebraic multiplicity) and s=2 (geometric multiplicity): $$\dot X(t) = AX(t)$$, Where A=

$$\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ -2 & -2 & -1 \end{bmatrix}$$

When I solve det(A−λ)= 0 I get 1 eigenvalue λ =1 with k=3, then I get 2 eigenvectors $$\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$$ $$\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}$$

Now I am lost, I really dont know what to do next or how to find the general form of X(t).

EDIT:

I found what to do in the case of k=3 and s=1: you solve (A−λ)$$V_1$$=0, (A−λ)$$V_2$$=$$V_1$$ and (A−λ)$$V_3$$ = $$V_2$$.

Then $$X(t)=C_1V_1e^{λ_1}+c_2e^{λ_1}(V_1t+V_2)+ C_3e^{λ_1}(\frac{V_1t^2}{2}+V_2t+V_3)$$.

I know that a part of the solution I'm looking is $$C_1V_1e^{λ_1}+ C_2V_2e^{λ_1}$$, with $$V_1$$ and $$V_2$$ the eigenvetors that I already have found.

So I am looking for the last part of the solution.

• The simplest way is to capitalize on the fact that $A-I$ is nilpotent. More conventionally, compute the Jordan normal form of $A$ and go from there, or somewhat less conventionally, use the Cayley-Hamilton theorem to write $e^{tA}=aI+bA+cA^2$ and solve for the unknown coefficients. – amd Nov 29 '19 at 0:31

## 2 Answers

When a matrix $$A$$ has only one eigenvalue $$\lambda$$ like this (and it’s not a multiple of the identity), then $$N=A-\lambda I$$ is nilpotent. Moreover, $$\lambda I$$ and $$N$$ commute, therefore $$e^{tA}=e^{t(\lambda I+N)}=e^{\lambda tI}e^{tN}=e^{\lambda t}e^{tN}$$. Since $$N$$ is nilpotent, the power series for $$e^{tN}$$ only has a finite number of terms. In this case, because the geometric multiplicity of the eigenvalue is two, you know that $$N^2=0$$, so that $$e^{tN}=I+tN$$.

Putting this all together, the solution to the differential equation is $$X(t) = e^t(I+t(A-I))X(0).$$ If you’re not given initial conditions, then $$X(0)$$ is simply three arbitrary constants.

You can use the same method regardless of the geometric multiplicity of the lone eigenvalue. If it were $$1$$ instead, the only difference would be that $$e^{tN}$$ has one more term since $$(A-\lambda I)^2\ne0$$.

• I think that a lambda is missing in the final answer +1 for nice answer amd – Aryadeva Nov 29 '19 at 1:17
• Thanks, this was very helpful. But what if I have 2 eigen values with multiplicity higher than 1? Can I use this method? – Jose Avalos Nov 29 '19 at 3:07

When the matrix A is 3 by 3, and the geometrical multiplicity is 2 and algebraic multiplicity is 3 the general solution is: $$\vec x(t)=c_1\vec v_1* \exp(\lambda t)+c_2\vec v_2*\exp(\lambda*t)+c_3\exp(\lambda*t)[t(\alpha\vec v_1 +\beta \vec v_2)+\vec u]$$ When: $$(*)\space(A-\lambda*I)vec(u)=\alpha\vec v_1 +\beta \vec v_2$$ $$\alpha$$ and $$\beta$$ are determined along the solution, as when you're solving $$(*)$$ you will have to set a condition which will assure the solution. Example: $$\left[ \begin{array}{ccc|c} 1&1&1 &\alpha\\ 0&0&0 & \alpha - \beta\\ 0 & 0 &0 & \alpha - \beta \end{array} \right]$$ The condition for a solutoin is: $$\space \alpha - \beta = 0 \space$$