How can I find the eigenvalues of $\alpha$? The question is given below:



My trial is:
Let $\lambda $ be an eigenvalue of $\alpha,$ then 
 $$ \alpha (
  \left[ {\begin{array}{cc}
   a \\
   b \\
   c \\
   d \\
  \end{array} } \right] 
) = \lambda
  \left[ {\begin{array}{cc}
   a \\
   b \\
   c \\
   d \\
  \end{array} } \right]
 = 
  \left[ {\begin{array}{cc}
   \lambda a \\
   \lambda b \\
   \lambda c \\
   \lambda d
  \end{array} } \right]
 $$
$$i.e. \left[ {\begin{array}{cc}
   b + c \\
   c \\
   0 \\
   0 \\
  \end{array} } \right] = \left[ {\begin{array}{cc}
   \lambda a \\
   \lambda b \\
   \lambda c \\
   \lambda d
  \end{array} } \right]
 \rightarrow (1.)$$
Now, for $\left[ {\begin{array}{cc}
   a \\
   b \\
   c \\
   d \\
  \end{array} } \right] $ to be an eigenvector, at least one of $a, b, c, d$ must be nonzero. WLOG, let $a \neq 0.$ Therefore from $(1.)$ we have $$\lambda = \frac {b + c}{a}, \lambda b = c \quad i.e. \quad  b^2 = c(a-b), \lambda c = 0 \quad  i.e. c^2 = (-bc).$$ but then what?
Could anyone help me in solving this, please? 
 A: From (1.), we have $\lambda c=0$ and $\lambda d=0$, so $\lambda=0$ is one solution.
Otherwise, $c=0$ and $d=0$, and $c=\lambda b$, so $\lambda=0$ or $b=0$.
But if $c=d=b=0$, then $\lambda a=b+c=0$, so $a=0$ too, so there are no eigenvalues besides $\lambda=0$.
A: Another way to do:
$$\alpha(a,b,c,d)=(b+c,c,0,0).$$
Assuming that $e_1,\dots, e_4$ is the canonical basis of $\mathbb{R}^4$. Then,
$$\alpha(e_1)=(0,0,0,0)$$
$$\alpha(e_2)=(1,0,0,0)$$
$$\alpha(e_3)=(1,1,0,0)$$
$$\alpha(e_4)=(0,0,0,0)$$
The matrix associated to $\alpha$ is:
$$[\alpha]=\begin{pmatrix}
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$$
Then, the charascterisc polynomail of $[\alpha]$ is
$$p(\lambda)= \det(\lambda I - [\alpha])= \det\left[\begin{pmatrix}
\lambda & 0 & 0 & 0 \\
0 & \lambda & 0 & 0 \\
0 & 0 & \lambda & 0 \\
0 & 0 & 0 & \lambda 
\end{pmatrix} - \begin{pmatrix}
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}\right] = \det\left[\begin{pmatrix}
\lambda & -1 & -1 & 0 \\
0 & \lambda & -1 & 0 \\
0 & 0 & \lambda & 0 \\
0 & 0 & 0 & \lambda 
\end{pmatrix}\right]$$
Can you proceed from here?
