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I want to find the eigenvectors of the following matrix A and given an eigenvalue of -1 with an algebraic multiplicity of 3.

$$ \begin{bmatrix} -1 & -1 & 0 \\ -1 & 0 & -1 \\ -1 & 2 & -2 \\ \end{bmatrix} $$ The formula I am using is $e^{tA} = e^{\lambda t(A- \lambda I)}$. I was able to find the first eigenvector and the first generalized eigenvector just fine, $\vec{v_1} =e^{-t}\begin{bmatrix} 1 \\ 0 \\ -1 \\ \end{bmatrix} $ and $\vec{v_2} = e^{-t}\begin{bmatrix} 1-t \\ 1 \\ t \\ \end{bmatrix} $ however, I encountered a problem. $(A- \lambda I)^3 $ is equal to

$$ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ So the series truncates before I am able to find the 2nd generalized eigenvector and I am uncertain how to proceed. Do I find the nullspace of the zero matrix then pick a basis vector that isn't a multiple of the previous two?

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  • $\begingroup$ Your generalized eigenvectors are of the form $s\cdot\pmatrix{1\cr 1\cr 0\cr} + t\cdot\pmatrix{-1\cr 0\cr 1\cr}$; that means you have more than one linearly independent eigenvector. So you won't get any more vectors by looking at the null space of $(A-\lambda I)^3$. $\endgroup$ – Christopher Carl Heckman May 9 '16 at 22:42
  • $\begingroup$ So my last vector is a linear combination of the first two? $\endgroup$ – Diehardwalnut May 9 '16 at 22:46
  • $\begingroup$ You get two different (linearly independent) generalized eigenvectors. $\endgroup$ – Christopher Carl Heckman May 9 '16 at 22:51

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