Say $ A\in\mathbb{C}^{6\times6} $ and has eigenvalues $\lambda_1$ and $\lambda_2$ of multiplicity $ 3$ both of them. And for $\kappa=1,2,3$ the echelon form of the matrix

$$ (A-\lambda_1I)^\kappa $$

has exactly $\kappa$ zero lines and for $\kappa=1,2$

$$ (A-\lambda_2I)^{\kappa} $$

has exactly $\kappa+1$ zero lines.

Find the Jordan of the matrix subject to $\lambda_1$, $\lambda_2$. What I did is for $\kappa=1$

$$ (A-\lambda_1I) $$

has $1$ zero line so the rank of the the matrix is $6-1=5$. And so the nullspace has rank of $1$. Thus the geometric multiplicity of $\lambda_1$ is $1$, so I have $1$ Jordan block with $\lambda_1$ same logic for $\lambda_2$ found $2$ Jordan blocks. So I have 1 Jordan block of $3\times3$ with $\lambda_1$ and one $1\times 1 $,$2\times 2$ for $\lambda_2$.

Question: what am I supposed to do with the rest information for the different values of $\kappa$?

  • $\begingroup$ Your argument about the matrix ranks is not $100\%$ true. For example: a matrix $\left[\matrix{1 & 1 & 1\\ 1 & 1 & 1\\ 0 & 0 & 0}\right]$ has exactly one zero line, but the rank is not $3-1=2$. All you can say is that with one zero line you have rank$\le 2$. $\endgroup$ – A.Γ. Aug 20 '15 at 18:15
  • $\begingroup$ It says it is in an echelon form (scalar form) $\endgroup$ – Manolis Lyviakis Aug 20 '15 at 18:17
  • $\begingroup$ Okay, then in this case you do not need more information. One block for $\lambda_1$ and two blocks for $\lambda_2$ makes the Jordan form unique in case of multiplicity $3$. $\endgroup$ – A.Γ. Aug 20 '15 at 18:23
  • $\begingroup$ So if it is like im saying it is , it is right ? $\endgroup$ – Manolis Lyviakis Aug 20 '15 at 18:25
  • 1
    $\begingroup$ Yes, it is right. $\endgroup$ – A.Γ. Aug 20 '15 at 18:33

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