How can I find the eigenvalues of $\alpha$?

Question

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?

Answer 1

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 $\lambda=0$.

Answer 2

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?