I'm having troubles finding the right base, I keep getting a weird Jordan form.
I need to find the Jordan form and basis of $$ \left[\begin{array}{cc}
1 & 1\\
-1 & 3
\end{array}\right] $$
this leads me to $$ \left[\begin{array}{cc}
\lambda-1 & -1\\
1 & \lambda-3
\end{array}\right]=(\lambda-1)(\lambda-3)+1=(\lambda-2)^{2} $$
So now I calculate the eigenvectors for eigenvalue 2:
$$ \begin{array}{cc|c}
1 & -1 & 0\\
1 & -1 & 0
\end{array}\Rightarrow\begin{cases}
x-y=0\\
x-y=0
\end{cases}\Rightarrow\left(\begin{array}{c}
1\\
1
\end{array}\right) $$
Now after I found it I find the eigenvalue that chains to it and get : $$ \begin{array}{cc|c}
1 & -1 & 1\\
1 & -1 & 1
\end{array}\Rightarrow\begin{cases}
x-y=1\\
x-y=1
\end{cases}\Rightarrow\left(\begin{array}{c}
1\\
0
\end{array}\right) $$
So now I have P=$$ \left(\begin{array}{cc}
1 & 1\\
1 & 0
\end{array}\right)\Rightarrow P^{-1}=\left(\begin{array}{cc}
0 & 1\\
1 & -1
\end{array}\right) $$
After multiplying
$$P^{-1}AP=\begin{pmatrix}0&1\\ \:\:1&-1\end{pmatrix}\begin{pmatrix}1&1\\ \:\:-1&3\end{pmatrix}\begin{pmatrix}1&1\\ \:\:1&0\end{pmatrix}=\left(\begin{array}{cc}
2 & -1\\
0 & 2
\end{array}\right) $$
Where did I go wrong? why is the Jordan form not $$ \left(\begin{array}{cc} 2 & 1\\ 0 & 2 \end{array}\right)? $$