Find a Jordan basis for the endomorphism $g:M_2(R)\longrightarrow M_2(R)$ such that... 
Find a Jordan basis for the endomorphism $g:M_2(R)\longrightarrow M_2(R)$ such that
$M(g,B) = \begin{pmatrix} 2&0&3&0\\ 1&2&0&3\\0&0&2&0\\ 0&0&1&2 \end{pmatrix}$, where $B=(\begin{pmatrix} 0&1\\ 1&1 \end{pmatrix},\begin{pmatrix} 1&0\\ 1&1 \end{pmatrix},\begin{pmatrix} 1&1\\ 0&1 \end{pmatrix},\begin{pmatrix} 1&1\\ 1&0 \end{pmatrix})$

I know how to calculate a Jordan basis. However, I am stuck at this problem, I really don't know how to start.
Thank you

Let $\alpha=\{E_1,E_2,E_3,E_4\}$ where
\begin{align*}
E_1 &=
\left[\begin{array}{rr}
0 & 1 \\
1 & 1
\end{array}\right]
&
E_2 &= 
\left[\begin{array}{rr}
1 & 0 \\
1 & 1
\end{array}\right]
&
E_3 &=
\left[\begin{array}{rr}
1 & 1 \\
0 & 1
\end{array}\right]
&
E_4 &=
\left[\begin{array}{rr}
1 & 1 \\
1 & 0
\end{array}\right]
\end{align*}
You are given that 
$$
[g]_\alpha^\alpha=
\left[\begin{array}{rrrr}
2 & 0 & 3 & 0 \\
1 & 2 & 0 & 3 \\
0 & 0 & 2 & 0 \\
0 & 0 & 1 & 2
\end{array}\right]
$$ and you wish to find a basis $\beta=\{F_1,F_2,F_3,F_4\}$ such that $[g]_\beta^\beta$ is in Jordan form. 
You state that you know how to find Jordan bases so I'll assume that you can show that $[g]_\alpha^\alpha=PJP^{-1}$ where
\begin{align*}
P &=
\left[\begin{array}{rrrr}
0 & 3 & 0 & 1 \\
6 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & -\frac{1}{3}
\end{array}\right]
&
J &=
\left[\begin{array}{rrrr}
2 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2
\end{array}\right]
\end{align*}
We can interpret $P$ as the change of basis matrix from $\alpha$ to $\beta$. This means
$$
[g]_\alpha^\alpha = \underbrace{[\DeclareMathOperator{id}{id}\id]_\beta^\alpha}_{=P}\underbrace{[g]_\beta^\beta}_{=J}\underbrace{[\id]_\alpha^\beta}_{=P^{-1}}
$$
Since
$$
P^{-1}=
\left[\begin{array}{rrrr}
0 & \frac{1}{6} & 0 & 0 \\
\frac{1}{6} & 0 & 0 & \frac{1}{2} \\
0 & 0 & 1 & 0 \\
\frac{1}{2} & 0 & 0 & -\frac{3}{2}
\end{array}\right]
$$
it follows that 
\begin{array}{rcrcrcrcrcrcrcrcrc}
F_1 &=& 0\cdot E_1 &+& \frac{1}{6}\cdot E_2 &+& 0\cdot E_3 &+& \frac{1}{2}\cdot E_4\\
F_2 &=& \frac{1}{6}\cdot E_1 &+& 0\cdot E_2 &+& 0\cdot E_3 &+& 0\cdot E_4\\
F_3 &=& 0\cdot E_1 &+& 0\cdot E_2 &+& 1\cdot E_3 &+& 0\cdot E_4\\
F_4 &=& 0\cdot E_1 &+& \frac{1}{2}\cdot E_2 &+& 0\cdot E_3 &-& \frac{3}{2}\cdot E_4\\
\end{array}
Your desired basis is thus
\begin{align*}
F_1 &=
\left[\begin{array}{rr}
\frac{2}{3} & \frac{1}{2} \\
\frac{2}{3} & \frac{1}{6}
\end{array}\right]
&
F_2 &=
\left[\begin{array}{rr}
0 & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6}
\end{array}\right]
&
F_3 &=
\left[\begin{array}{rr}
1 & 1 \\
0 & 1
\end{array}\right]
&
F_4 &=
\left[\begin{array}{rr}
-1 & -\frac{3}{2} \\
-1 & \frac{1}{2}
\end{array}\right]
\end{align*}
