Stack Exchange
Finding Jordan Basis of a matrix (3X3)
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?

Are you claiming $\begin{pmatrix} 0\ 1\ 0 \end{pmatrix}$ is an eigenvector of the eigenvalue $1$?

@GitGud I think so...

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.

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

Ok, and now how can I find the jordan base (and hence P)?

Well, you already can choose two lin. independent vectors as shown above, and the third one shall be a generalized eigenvector, as already told...
