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If I'm solving a linear system $\vec{x}^{\,'}=A\vec{x}$ with constant square matrix $A$ and if $A$ has an eigenvalue $\lambda$ with multiplicity $3$ but there's only 2 eigenvectors $\vec{v_1},\vec{v_2}$ corresponding to $\lambda$. When finding a third linearly independent eigenvectors $\vec{v_3}$, which equation should I be solving? $(A-\lambda I)\vec{v_3}=\vec{v_1},\vec{v_2}$ or $\vec{v_1}+\vec{v_2}?$

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When the matrix is defective you need to find linearly independent generalised eigenvectors, i.e. vectors in the kernel of $(A - \lambda I)^{n}$.

Generalised eigenvectors, as you suggest in your question, can be found by looking for a chain that satisfies $(A-\lambda I)\textbf{w}_{k} = \textbf{w}_{k-1}$ and $(A - \lambda I) \textbf{w}_{1} = \textbf{0}$. Here $\textbf{w}_{1}$ is a regular old eigenvector.

In the case you describe (algebraic multiplicity 3, geometric multiplicity 2) not all the eigenvectors of $\lambda$ start a non-trivial chain. You will need to solve the equation

$$ (A-\lambda I)\textbf{w} = \alpha\textbf{v}_{1} + \beta\textbf{v}_{2} $$ for some suitably chosen constants $\alpha$ and $\beta$. If you find it hard to choose appropriate constants you can calculate the generalised eigenvector directly by calculating the matrix $(A - \lambda I)^{2}$ and solving $$ (A-\lambda I)^{2}\textbf{w} = \textbf{0}.$$

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  • $\begingroup$ Instead of using the linear combination, can I find new vectors by solving $(A-\lambda I)v_3=v_1$ or $(A-\lambda I)v_3=v_2$? Or should I never bother using this approach? $\endgroup$ – Iced Palmer Nov 14 '17 at 8:46
  • $\begingroup$ @IcedPalmer There is no guarantee that you can find a solution to those equations. Not every eigenvector lies at the base of a non-trivial chain of generalised eigenvectors. $\endgroup$ – in_mathematica_we_trust Nov 14 '17 at 9:15

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