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I have a 3*3 matrix A= $$\begin{pmatrix} 3 & 0 & -1 \\ -1 & 2 & 1 \\ 1 & 2 & 3 \\ \end{pmatrix}$$ that has two eigenvalues $\lambda_1$=$\lambda_2$=2 and $\lambda_3$=4 so the eigenvalue which equals 2 has algebraic multiplicity two. However the eigenvector which equals 2 only has one vector which spans its eigenspace and therefore has geometric multiplicity 1. As usual there is one vector spanning the eigenspace corresponding to eigenvalue 4. However when I look for generalised eigenvectors corresponding to eigenvalue 2 by working out ker((A-2I)^2), only the zero vector is contained in it. How do I find a basis of eigenvectors and generalised eigenvecotrs of the matrix A if the only generalised eigenvector is the zero vector, which by definition is not allowed?

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  • $\begingroup$ The vector $(1,0,0)$ is in $\ker\left(\left(A-2I_3\right)^2\right)$. $\endgroup$ – Git Gud Mar 23 '15 at 23:11
  • $\begingroup$ (1,0,0) is not in ker((A−2I)^2) as (A-2I)^2= $$ \begin{matrix} 0 & -2 & 0 \\ 2 & 2 & 2 \\ -2 & 2 & 2 \\ \end{matrix} $$ $\endgroup$ – user3544587 Mar 23 '15 at 23:29
  • $\begingroup$ Using MATLAB's jordan command, I find a generalized eigenvector of $(1,-1/2,1/2)$. Checking, it is indeed in the kernel of $(A-2I)^2$. I also find that you have $(A-2I)^2$ wrong, and that its first column is zero (so that as Git Gud suggested, $(1,0,0)$ should also be a generalized eigenvector). $\endgroup$ – Ian Mar 23 '15 at 23:43
  • $\begingroup$ @user3544587 Just by looking one can tell that the matrix you mentioned in your comment is invertible (first and last column are orthogonal), contradicting the meaning of eigenvalues, so you got that one wrong. $\endgroup$ – Git Gud Mar 24 '15 at 0:06
  • $\begingroup$ Oh yes, I'm sorry guys I realised I made a calculation error. Thank you for the help, really appreciate it. $\endgroup$ – user3544587 Mar 24 '15 at 0:15

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