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I have a matrix:

$$A=\begin{pmatrix}0&1&0\\-4&4&0\\-2&1&2\end{pmatrix}$$

solving $\det|A-\lambda{I}|$ I got characteristic polynom that equals to $(2-\lambda)^3 = 0$ for eigenvalue found two eigenvectors and one generalized eigenvector: $v_1=(1,2,0)\quad v_2=(0,0,1) \quad v_3=(1,0,0)$

What do I have to do to find Jordan basis here? (and what do I need to find Jordan basis in general, I mean is there appropriate alghoritm?, What I read did not make things more clear).

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  • $\begingroup$ The eigenvectors and generalized eigenvectors, together, form the Jordan basis. But I think your first eigenvector is wrong. $\endgroup$ – Ian Sep 17 '17 at 21:49
  • $\begingroup$ The characteristic polynomial of this matrix is not $(2-\lambda)^3$. $\endgroup$ – Bernard Sep 17 '17 at 22:01
  • $\begingroup$ @Bernard, I confused one sign in the matrix, should be fixed now $\endgroup$ – M.Mass Sep 17 '17 at 22:04
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Here the way to go: consider the sequence of kernels: $$\{\,0\,\}\varsubsetneq\ker(A-2I)\varsubsetneq\ker(A-2I)^2\subset\dots$$ The sequence stops after step $2$ since $$A-2I=\begin{bmatrix}-2&1&0\\-4&2&0\\-2&1&0\end{bmatrix}\qquad (A-2I)^2=\begin{bmatrix} 0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ $A-2I$ has rank $1$, hence its kernel (the eigenspace) has codimension $1$, i.e. has dimension $2$.

$(A-2I)^2$ is the null matrix, hence its kernel has dimension $3$. Take any vector in $\ker(A-2I)^2\smallsetminus\ker(A-2I)$, i.e. any vector of $\mathbf R^3$ which is not an eigenvector. As the eigenspace is defined by the equation $\; y=2x$, we'll take, say $$e_3=(0,1,0). $$ Note $e'_2=(A-2I)e'_3=(1,2,1),\;$ is an eigenvector by construction1,2,0. We complete this set of two vectors to a basis, by choosing another eigenvector, linearly independent from $e'_2$, say $$e'_1=(1,2,0).$$ The definition of $e'_2$ from $e'_3$ can be written as $\; Ae'_3=2e'_3+e'_2$, so the matrix of the lineap map in basis $(e'_1,e'_2,e'_3)$ is the Jordan form: $$J=\begin{bmatrix}2&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$

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