Finding $[T_{|W_i}]_{C_i}$ Let $B=\{v_1,v_2,v_3\}$, a basis of $V$ above $\mathbb{R}$. Let $$ [T]_B = \left(\begin{array}{cccc} 6&-3&-2\\4&-1&-2\\10&-5&-3 \end{array}\right)$$
The characteristic polynomial is $f_T(x) = (x-2)(x^2+1)$. Hence, $m(x) = (x-2)(x^2+1)$
We have $W_1 = \ker (T-2I) = \text{span}\{v_1,2v_3\}$ and $W_2 = \ker (T^2+I) = \text{span}\{v_1+v_2, v_3\}$ and $V = W_1 \oplus W_2$.
Now, we denote $$C_1 = \{v_1 + 2v_3\} \\ C_2 = \{ v_1+v_2,v_3 \}$$
I don't quite understand why 
$$[T_{|W_1}]_{C_1} = (2) \\ [T_{|W_2}]_{C_2} = \left(\begin{array}{cccc} 3&-2\\5&-3 \end{array}\right)$$

$ \{v_1 + 2v_3\}=\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\\
\{v_1+v_2,v_3\}  =\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right),\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)$ 
and $T\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\ = 2\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\\$ 
thus your first matrix comes out as $(2)$
Now, $T\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right)=3\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right)+5\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)$
$T\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)=-2\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right)  -3\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)$ 
hence second matrix comes out as $\left(\begin{array}{cccc} 3&-2\\5&-3 \end{array}\right)$
