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How to solve ${\bf x}' = A{\bf x}$ with $A=\begin{pmatrix} -3 & 2 \\ -2 & 1 \\ \end{pmatrix}$?

Well, so first I got the eigenvalues which both are $-1$, and eigen vector is $(1,1)$. Then I wrote the answer as ${\bf x}(t) = c_1e^{-t}(1,1) + c_2e^{-t}(2,2)$

In the answer It's written like that:

$x = (C_1+2C_2t)e^{-t}$

$y = (C_1+C_2+2C_2t)e^{-t}$

Can someone explain please how they got to that answer?

Edit: Like I written, I can show that $x = (C_1+2C_2t)e^{-t}$, but when I'm trying to show $y$, I can't add the $+C_2$ there.

Ilan Aizelman WS
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  • Related: http://math.stackexchange.com/q/441121/9464 –  Dec 23 '16 at 14:31
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    @Moo, defective might be the word that you're looking for. The Jordan form will certainly be helpful here. – J. M. ain't a mathematician Dec 23 '16 at 14:32
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    Your solution for ${\bf x}(t)$ is incorrect.

    There is a systematic way to deal with the ODE ${\bf x}'=A{\bf x}$ when $A$ is not diagonalizable. Any "reasonable" ODE textbook should contain a section about this topic.

    You might want to try first a "simpler" version of your problem:

    ${\bf x}'=A{\bf x}$ where $A= \begin{pmatrix} 1&1\0&1 \end{pmatrix}. $

     –  Dec 23 '16 at 14:46
  • @Jack I don't know myself why they gave me a matrix that isn't diagonalizable – Ilan Aizelman WS Dec 23 '16 at 17:06
  • What book are you reading? What is the context for this exercise? –  Dec 23 '16 at 17:09
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    "I don't know myself why they gave me a matrix that isn't diagonalizable" Because "they" wanted you to learn the method to solve the non diagonalizable case (and this step should be over) and to apply this method to the present example, obviously. – Did Dec 23 '16 at 18:49

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When you have a repeated eigenvalue of multiplicity $n$ but fewer than $n$ associated eigenvectors, then the eigenvalue is called "defective." That's the case here, with $n=2$. If $v$ is the eigenvector, then one solution to the system is $e^{\lambda t}v$. But you need a second, linearly independent solution and the one you give is not linearly independent (because $(2,2)$ is a scalar multiple of $(1,1)$.

A second, linearly independent solution is given by $e^{\lambda t}(tv+w)$ where $w$ is a "generalized" eigenvector. It's the solution to $(A-\lambda I)w=v$. (Actually, it's the solution to $(A-\lambda I)^2 w = 0.$ So in your case, solve

$$\begin{pmatrix} -2 & 2 \\ -2 & 2 \\ \end{pmatrix} w =\begin{pmatrix} 2\\2\\ \end{pmatrix}$$

(where I choose the second version of the eigenvector to match the solution you were given.)

Take $w = (0,1)$ and you have the solution you were given.

B. Goddard
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