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Having trouble finding the jordan base for this matrix \begin{pmatrix} 1 & 1 &0 \\ 0 &1 &1 \\ 0& 0 &2 \end{pmatrix} I know that the Characteristic polynomial is : (t-1)^2(t-2) I started with eigenvalues λ=1 I found that the minimal k is 2 and: dim(ker(I-A))=...=Span({\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}} and- dim(ker(I-A)^2)=...=Span({\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}})

Which vector i have to choose for Jordan Basis? What is the next steps?

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    $\begingroup$ Are you claiming $\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}$ is an eigenvector of the eigenvalue $1$? $\endgroup$ – Git Gud Mar 31 '14 at 16:17
  • $\begingroup$ @GitGud I think so... $\endgroup$ – I need your help Mar 31 '14 at 16:22
  • $\begingroup$ It isn't. You did something wrong. note that if $A$ is your starting matrix, then $A-I_3$ has rank $2$, so you can only get $3-2\color{grey}{=1}$ linearly independent eigenvectors. $\endgroup$ – Git Gud Mar 31 '14 at 16:25
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$$\lambda I-A=\begin{pmatrix}\lambda-1&-1&0\\ 0&\lambda-1&-1\\ 0&0&\lambda-2\end{pmatrix}$$

so

$$\lambda=1:\;\;\begin{cases}-y=0\\ -z=0\\-z=0\end{cases}\implies \begin{pmatrix}x\\0\\0\end{pmatrix}\;,\;\;x\neq 0\,,\;\;\;\text{is an eigenvector for}\;\;\lambda =1$$

$$\lambda=2:\;\;\begin{cases}x-y=0\\ y-z=0\end{cases}\implies \begin{pmatrix}x\\x\\x\end{pmatrix}\;,\;x\neq0\,,\,\,\;\text{is an eigenvector for}\;\;\lambda =2$$

Take it from here (you only need one more generalized eigenvector...)

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  • $\begingroup$ Ok, and now how can I find the jordan base (and hence P)? $\endgroup$ – I need your help Mar 31 '14 at 16:33
  • $\begingroup$ Well, you already can choose two lin. independent vectors as shown above, and the third one shall be a generalized eigenvector, as already told... $\endgroup$ – DonAntonio Mar 31 '14 at 16:34

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