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?
$$\lambda I-A=\begin{pmatrix}\lambda-1&-1&0\\ 0&\lambda-1&-1\\ 0&0&\lambda-2\end{pmatrix}$$
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...)