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M y question is relating to the matrix as A. I have started off this problem by finding the eigenvalues, which turns out to be 3 ( I should note that it has an algebraic multiplicty of 3)

From there I have found the corresponding Eigenspace which is $E_3=span(-1,1,1)$

I am little confused as to what to do from here.

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The Jordan form can be deduced from the information you already have, namely that the eigenspace is 1-dimensional and all of the eigenvalues are 3. The Jordan matrix is, \begin{pmatrix} 3 & 1 & 0\\ 0 & 3 & 1\\ 0 & 0 & 3 \end{pmatrix} where the eigenvalues are on the diagonal, and the size of the Jordan block is 3 by 3, due to the dimension of the eigenspace.

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  • $\begingroup$ Isn't the dimension of my Eigenspace 1? so how did you get a 3 by 3 matrix $\endgroup$ – HueHue Jun 20 '16 at 13:03
  • $\begingroup$ If the eigenspace was 3D, then you could construct an orthonormal basis consisting of eigenvectors and you'd get a diagonal matrix, or a matrix with 3 $1 \times 1$ Jordan blocks on the diagonal. As the geometric multiplicity becomes less than the algebraic multiplicity, the size of the Jordan block increases because you can't construct an orthogonal basis of eigenvectors for the eigenspace. Hope this clarified a bit. $\endgroup$ – Merkh Jun 20 '16 at 13:07

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