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I have the following system:

$$\begin{cases} x'= 2x + 2y -3z \\ y' = 5x + 1y -5z \\ z' = -3x + 4y \end{cases} $$

$$\det(A - \lambda I)= -(\lambda - 1)^3$$

the eigenvector for my single eigenvalue is : $v_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

I can't find any other linearly independent solutions.

What can you do in this $3\times 3$ case with only a single eigenvalue?

If you wish you can provide a reference and then I will try my hand at solving it myself.

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    $\begingroup$ Hint: Generalised Eigenvectors. As a reference, I asked the same question some time ago: math.stackexchange.com/questions/825804/… $\endgroup$ – Nigel Overmars Oct 22 '14 at 15:18
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    $\begingroup$ Method 1, Method 2. $\endgroup$ – Git Gud Oct 22 '14 at 15:19
  • $\begingroup$ @Amzoti Did I correctly use method 1 linked by Git Gud? $\endgroup$ – Monolite Oct 24 '14 at 14:24
  • $\begingroup$ Yes I notice now too, I will redo the calculations in matlab. I am happy the process is correct. $\endgroup$ – Monolite Oct 24 '14 at 15:04
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Suppose $n \times n$ matrix $A$ has only a single eigenvector $\lambda$, and let $B = A - \lambda I$. Then $B^n = 0$. So $$\exp(A t) = \exp(\lambda t) \exp(B t) = \exp(\lambda t) \sum_{j=0}^{n-1} B^j t^j/j!$$ and the solutions of $x' = A x$ are linear combinations of the columns of this matrix. In this case $n=3$, $$B = A - I = \pmatrix{1 & 2 & -3\cr 5 & 0 & -5\cr -3 & 4 & -1\cr}, \ B^2 = \pmatrix{20 & -10 & -10\cr 20 & -10 & 10\cr 20 & -10 & -10\cr}$$ and $$ \exp(At) = e^t \pmatrix{1 + t + 10 t^2 & 2t - 5 t^2 & -3t - 5 t^2\cr 5 t + 10 t^2 & 1 - 5 t^2 & -5 t + 5 t^2\cr -3t + 10 t^2 & 4 t - 5 t^2 & 1 - t - 5 t^2}$$

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  • $\begingroup$ Is the method I used also correct? $\endgroup$ – Monolite Oct 24 '14 at 10:32
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    $\begingroup$ @Monolite Yes, it is also correct. $\endgroup$ – Robert Israel Oct 24 '14 at 14:47
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Following: Method 1 linked by Git Gud and suggested by Amzoti.

I Now know that a generalized eigenvector can be found by:

$$(A-\lambda I)v_2=v_{1}$$

I obtain $ v_2 = \begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix}$ so an other solution is:

$$x_2(t) =e^t \left( t \begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix} + \begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix} \right) $$

But I need three solutions so I must find a $v_3$, as before:

$$(A-\lambda I)v_3=v_{2}$$

I now obtain $v_3 = \begin{bmatrix} 0.08 \\ 0.4\\ 0\end{bmatrix}$ and the last solution needed is: $$ x_3(t) = e^{ t}\left(\dfrac{t^2}{2!}\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}+t\begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix}+ \begin{bmatrix} 0.08 \\ 0.4\\ 0\end{bmatrix} \right)$$

So my general solution would look like:

$$X(t) = C_1e^t\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}+ C_2 e^t \left( t \begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix} + \begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix} \right) + C_3e^{ t}\left(\dfrac{t^2}{2!}\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}+t\begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix}+ \begin{bmatrix} 0.08 \\ 0.4\\ 0\end{bmatrix} \right) $$

Correct?

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When you have a triple eigenvalue and 1 single eigenvector, you have to find two eigenvectors $\mu, \text{ and }\rho,$which satisfy $$(A-\lambda I)\vec{\rho} = \vec{\eta}$$ $$\text{and}$$ $$(A-\lambda I)\vec{\mu} = \vec{\rho}.$$Then, the solution will be $$\frac{1}{2}t^2e^{\lambda t}\vec{\eta}+te^{\lambda t}\vec{\rho}+e^{\lambda t}\vec{\mu}.$$

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