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I have a 3x3 matrix A with the following elements:

$$\begin{pmatrix} 5 & -4 & 2 \\ 4 & -5 & 4 \\ 6 & -12 & 9 \end{pmatrix}$$

The ODE is $y' = Ay$ with $y(0) = [1 \quad 0 \quad 1]^T$.

I found the eigenvalue to be 3 with a multiplicity of 3. I found the two eigenvectors to be:$[2 \quad 1 \quad 0]^T$ and $[-1 \quad 0 \quad 1]^T$ (both of these transpose). I feel like I need a 3rd eigenvalue to solve this ODE but not sure what to do. Also not sure how to type matrices so sorry for confusion

Tanny Sieben
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  • Hello and welcome to MSE! (I'm fairly new myself.). Notes and links for how to format are here: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. I think they call it "array" – RobertTheTutor Mar 18 '21 at 23:12
  • I tried $\pmatrix{0 & 1 & a}^\top$ and found its an eigenvector when $a=2$. The other eigenvectors have zero on the 2nd and 3rd elements, so I decided to find one with 0 in the first element. – John Alexiou Mar 18 '21 at 23:57
  • So the solution is the superposition of three modes with this form $$y = {\rm e}^{3x} \sum_{i=1}^3 (a_i + b_i x) $$ where the $b_i$ coefficients are the steady state solution and $a_i$ dependent on the initial conditions. – John Alexiou Mar 19 '21 at 03:28
  • In short: $3$ is an eigenvalue, but of multiplicity $2$, not $3$. There is another eigenvalue/eigenvector to get a full basis of eigenvectors. – hardmath Mar 19 '21 at 22:07
  • characteristic eq for A is -(lambda -3)^3 so 3 has a multiplicity of 3 –  Mar 19 '21 at 22:28

1 Answers1

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The third vector is called a generalized eigenvector. Instead of $(A-\lambda I)v = 0$, we use $(A - \lambda I)^2 v = 0$. In particular, given an eigenvector $v_1$, you usually want $(A - \lambda I) v_2 = v_1$. This can be repeated if you need more.

RobertTheTutor
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  • I keep lambda as lambda or substitute 3 for lambda since it's the eigenvalue? I've also found that (A-3I)^2 is the 0 matrix –  Mar 19 '21 at 21:56