Find the eingvalues of a Linear Transformation Let $V$ be a real vector space and $\{v_1,v_2, \cdots, v_n\}$  a basis of $V$. If $T:V \mapsto V$ is a linear transformation such that $T(v_1)=T(v_n)=v_1+v_n$ and $T(v_i)=v_i$ for all $i \in \{2,3, \cdots, n-1\}$, prove that the eingvalues of $T$ are 0,1,2.
I can show that 
$T(v_1 - v_n)=0$. Does it proves that $0$ is an eingvalue?
In the same approach we can show that $T(v_1 + v_n)=2(v_1 + v_n)$ but im not sure if this answer the question or i have to express the linear transformation in term of a matrix, so far i found that $T$ can be written as 
$A=\begin{pmatrix}
 1&  0&  \cdots&  0& 1\\ 
 0&  1&  0&  0&0 \\ 
 \vdots& 0 & 1 &  0& \vdots\\ 
 0&  0&  0&1  & 0\\ 
 1&  0&  \cdots&  0& 1
\end{pmatrix}$
but dont know how to finish from here.
 A: Recall that a scalar $\lambda$ is called an eigenvalue of a linear transformation $T:V\rightarrow V$ if there is a nonzero vector $v\in V$, called an eigenvector of T, such that $T(v)=\lambda v.$ So the answer to your first question

I can show that 
  $T(v_1 - v_n)=0$. Does it proves that $0$ is an eingvalue?

is "yes". For the vector $v_1 - v_n$ is nonzero. Similarly, the equations $T(v_1+v_n)=v_1+v_n$ $T(v_i)=v_i(i=2,\dots,n-1)$ imply that $1$ and $2$ are eigenvalues of $T$. 
But how can we be sure that T has no other eigenvalues? Set $$v_{i}'=\begin{cases}
v_{1}-v_{n} & \text{if }i=1\\
v_{i} & \text{if }i=2,\dots,n-1\\
v_{1}+v_{n} & \text{if }i=n
\end{cases}.$$ The set of vectors $\{v'_1, \dots, v'_n\}$ will form a basis of $V$, with respect to which $T$ has the matrix representation $$A'=\left[\begin{array}{ccccc}
0\\
 & 1\\
 &  & \ddots\\
 &  &  & 1\\
 &  &  &  & 2
\end{array}\right].$$ Suppose that $\lambda$ is an eigenvalue of $T$, that $v\in V$ is an eigenvector corresponding to $\lambda$, and that $x\in \mathbb{R} ^n$ is the coordinate vector of $v$ with respect to the basis $\{v'_1, \dots, v'_n\}$, (i.e. if $x=(x_1,\dots ,x_n)^T$, then $v'=x_1v'_1+\cdots +x_nv'_n)$. Then we have $A'x=\lambda x$. We can now turn to direct computation to see that $\lambda =0,1,$ or $2$.

But in this case, it might actually be quicker to just systematically compute the characteristic polynomial of $T$, for $\det(tI-A)$ ($A$ is the matrix that you found) can be computed fairly easily using row and column expansion.
