We let $v_{1}, v_{2}, v_{3}$ be the canonical basis of $\mathbf{R}^{3}$ and compute
$$
T\left(v_{1}\right)=T(1,0,0)=(1,1,0,-1),
$$
thus the first row of the matrix $A$ in the previous discussion is $(1,1,0,-1)$. We do the same with $v_{2}, v_{3}$ and we obtain
$$
A=\left[\begin{array}{cccc}
1 & 1 & 0 & -1 \
1 & -1 & 1 & 0 \
1 & 0 & -1 & 1
\end{array}\right]
$$
Using row-reduction we compute
$$
A_{r e f}=\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & -1
\end{array}\right]
$$
and we deduce that
$$
\operatorname{rank}(T)=3 .
$$

We need to prove that if $l$ is a linear form on $W$, then
$$
l \circ\left(T_{1}+c T_{2}\right)=l \circ T_{1}+c l \circ T_{2} .
$$
This follows from the fact that $l$ is linear.

a) Let $l$ be a linear form on $V_{3}$. Then
$$
\begin{gathered}
{ }^{t}\left(T_{2} \circ T_{1}\right)(l)=l \circ\left(T_{2} \circ T_{1}\right)=\left(l \circ T_{2}\right) \circ T_{1}= \
{ }^{t} T_{1}\left(l \circ T_{2}\right)={ }^{t} T_{1}\left({ }^{t} T_{2}(l)\right)={ }^{t} T_{1} \circ{ }^{t} T_{2}(l) .
\end{gathered}
$$
The result follows.
b) Since $T$ is an isomorphism, there is a linear transformation $T^{-1}$ such that $T \circ$ $T^{-1}=T^{-1} \circ T=$ id. Using part a) and the obvious equality ${ }^{t}$ id $=\mathrm{id}$, we obtain
$$
{ }^{t} T \circ{ }^{t}\left(T^{-1}\right)=\mathrm{id}={ }^{t}\left(T^{-1}\right) \circ t T,
$$
from where the result follows.

Let $x_{n}=\operatorname{Tr}\left(A^{n}\right)$. Multiplying relation (2.1) by $A^{n}$ and taking the trace yields
$$
x_{n+2}-\left(\lambda_{1}+\lambda_{2}\right) x_{n+1}+\lambda_{1} \lambda_{2} x_{n}=0 .
$$
Since $x_{0}=2$ and $x_{1}=\operatorname{Tr}(A)=\lambda_{1}+\lambda_{2}$, an immediate induction shows that $x_{n}=\lambda_{1}^{n}+\lambda_{2}^{n}$ for all $n$.

For the second part, let $z_{1}, z_{2}$ be the eigenvalues of $A^{n}$. By definition, they are the solutions of the equation $t^{2}-\operatorname{Tr}\left(A^{n}\right) t+\operatorname{det}\left(A^{n}\right)=0$. Since $\operatorname{det}\left(A^{n}\right)=(\operatorname{det} A)^{n}=$ $\lambda_{1}^{n} \lambda_{2}^{n}$ and $\operatorname{Tr}\left(A^{n}\right)=\lambda_{1}^{n}+\lambda_{2}^{n}$, the previous equation is equivalent to
$$
t^{2}-\left(\lambda_{1}^{n}+\lambda_{2}^{n}\right) t+\lambda_{1}^{n} \lambda_{2}^{n}=0 \quad \text { or } \quad\left(t-\lambda_{1}^{n}\right)\left(t-\lambda_{2}^{n}\right)=0 .
$$
The result follows.