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I was thinking about an ODE system which has three equal eigenvalues like this situation:

EDIT: $X'=AX$ $$A=\begin{pmatrix} 1 && 3 && 2 \\ 0 && 1 && 0\\ 0 && 2 && 1 \end{pmatrix} $$The eigenvalues are: $$(\lambda-1)^{3}=0$$ $$\lambda_{1,2,3}=1$$ It's clear that I have $\lambda=1$ with $\mu=3$.

I know very well the form of the solutions for a 2x2 system with eigenvalues with $\mu=2$:

$$ \begin{pmatrix} x(t)\\ y(t)\\ \end{pmatrix}=c_1e^{\lambda t}\mathbb u+c_2e^{\lambda t}[t\mathbb u+\mathbb w] $$

Now I'm asking myself which is a possible generalization for a 3x3 case of this formula. I'm expecting to have a $t^{2}$ term to appear in the formula.

Moreover, I have already computed the eigenvector plus two generalized eigenvectors (for the case I showed in this example for $\lambda=1$).

Eigenvectors: $$\mathbb u=(1,0,0)$$ $$\mathbb v=(0,0,\frac{1}{2})$$ $$\mathbb w=(0,\frac{1}{4},-\frac{3}{8})$$

Note that $\mathbb v$ and $\mathbb w$ are generalized eigenvectos.

Thank you very much.

muserock92
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  • Assuming you mean a homogeneous, first-order, linear system, since you have a guess, why not work out an example manually? Try ${\bf x}' = \pmatrix{\lambda&1&0\0&\lambda&1\0&0&\lambda} {\bf x}$. – Travis Willse Feb 24 '18 at 09:51
  • Thank you for your reply. I have updated the post with more details. I have omitted the calculus of the eigenvectors. Thank you :). – muserock92 Feb 24 '18 at 10:00
  • I finished multiplying it out. The common ground with this one: https://math.stackexchange.com/questions/2665177/what-is-the-mistake-in-my-matrix-power-formula/2665204#2665204 is that we can, with effort, find Jordan Form $J$ and $P^{-1} A P = J.$ However, the thing that can be calculated is some $f(J).$ Then we must reverse things, namely $P f(J) P^{-1} = f(A).$ This all requires keeping careful track of the matrices $P, P^{-1},$ not just a few eigenvectors/characteristic vectors – Will Jagy Feb 25 '18 at 18:51

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For any single Jordan block, the diagonal part and the nilpotent part can be regarded as commuting square matrices. When matrices $A,B$ commute, we get $e^{A+B} = e^A e^B = e^B e^A.$

You want $$e^{(I+N)t} = e^{tI} e^{tN} = e^t I e^{tN} = e^t e^{tN}$$ with $$ N = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) $$ and $$ N^2 = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) $$ and $$ N^3 = N^4 = N^5 = \cdots = 0 $$ So $$ e^{tN} = I + t N + \frac{t^2 }{2} N^2 = \left( \begin{array}{rrr} 1 & t & \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{array} \right) $$

$$ \frac{1}{8} \left( \begin{array}{rrr} 4 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 2 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & -3 \\ 0 & 0 & 4 \\ 0 & 8 & 0 \end{array} \right) = I $$

$$ \frac{1}{8} \left( \begin{array}{rrr} 4 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 2 & 0 \end{array} \right) \left( \begin{array}{rrr} 1 & 1& 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & -3 \\ 0 & 0 & 4 \\ 0 & 8 & 0 \end{array} \right) = \left( \begin{array}{rrr} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right) = A $$ $$ \frac{e^t}{8} \left( \begin{array}{rrr} 4 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 2 & 0 \end{array} \right) \left( \begin{array}{rrr} 1 & t& \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & -3 \\ 0 & 0 & 4 \\ 0 & 8 & 0 \end{array} \right) = e^{tA} $$ The denominators clear nicely, $$ e^{At} = \left( \begin{array}{ccc} 1 & 2 t^2 + 3 t & 2t \\ 0 & 1 & 0 \\ 0 & 2t & 1 \\ \end{array} \right) \; \; e^t \; \; = \; \; \left( \begin{array}{ccc} e^t & (2 t^2 + 3 t)e^t & 2t e^t \\ 0 & e^t & 0 \\ 0 & 2t e^t & e^t \\ \end{array} \right) $$

Will Jagy
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  • Thank you for your reply. I have pretty understood your computation, but it could be strange but unfortunately I still have some troubles rewriting that matrix in the form I needing for. Could you help me to reach the final results? Thank you very much. – muserock92 Feb 24 '18 at 21:58