The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} 1 \ { – 2} \ 0 \ 0 \end{array}} \right]}$.
An eigenvector of a matrix $A$ is a nonzero vector $v$ such that there exists a scalar $\lambda$, called the eigenvalue, such that $Av=\lambda v$.
To find the eigenvectors of a matrix, we can use the following steps:
- Find the characteristic polynomial of the matrix, which is the determinant of the matrix $A-\lambda I$.
- Solve the characteristic polynomial for $\lambda$.
- For each $\lambda$ that is a root of the characteristic polynomial, find a vector $v$ such that $Av=\lambda v$.
In this case, the characteristic polynomial of the matrix $A$ is
$$p(\lambda)=\det\left[ {\begin{array}{*{20}{c}} 5-\lambda & 0 & 0 & 0 \ 0 & 5-\lambda & 5 & 0 \ 0 & 0 & 2-\lambda & 1 \ 0 & 0 & 3 & 1-\lambda \end{array}} \right]$$
Expanding the determinant, we get
$$p(\lambda)=(5-\lambda)^4-20(5-\lambda)^3+100(5-\lambda)^2-200(5-\lambda)+250$$
Solving for $\lambda$, we find that the eigenvalues of the matrix $A$ are $\lambda=5,2,1,1$.
To find an eigenvector corresponding to the eigenvalue $\lambda=5$, we can use the following vector:
$$v=\left[ {\begin{array}{*{20}{c}} 1 \ { – 2} \ 0 \ 0 \end{array}} \right]$$
We can verify that $Av=\lambda v$ by substituting $v$ into the equation $Av=\lambda v$. We get
$$A\left[ {\begin{array}{{20}{c}} 1 \ { – 2} \ 0 \ 0 \end{array}} \right]=\left[ {\begin{array}{{20}{c}} 5 & 0 & 0 & 0 \ 0 & 5 & 5 & 0 \ 0 & 0 & 2 & 1 \ 0 & 0 & 3 & 1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1 \ { – 2} \ 0 \ 0 \end{array}} \right]=\left[ {\begin{array}{{20}{c}} 5 \ { – 10} \ 0 \ 0 \end{array}} \right]=\lambda\left[ {\begin{array}{{20}{c}} 1 \ { – 2} \ 0 \ 0 \end{array}} \right]=\left[ {\begin{array}{{20}{c}} 5 \ { – 10} \ 0 \ 0 \end{array}} \right]$$
Therefore, the vector $\left[ {\begin{array}{*{20}{c}} 1 \ { – 2} \ 0 \ 0 \end{array}} \right]$ is an eigenvector of the matrix $A$ corresponding to the eigenvalue $\lambda=5$.