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

Question

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.

Answer 1

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 $(A-\lambda I)^2\ne0$.

Answer 2

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: $\space \alpha - \beta = 0 \space$