Given a matrix $A$ and what it maps two vectors to, is $0$ an eigenvalue of it? Studying for my Algebra exam, and this question popped out with no solution in a previous exam:

Given a matrix $A$ such that $A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix},\ A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix}$.
(I) Is $0$ an eigenvalue of the matrix?
(II) Find a matrix like that, where the sum of its' eigenvalues is $0$.

So I (think) solved (I) but have no clue for (II).
Here's my solution for (I):

$A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = A \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix} = 0$, and then, the vector $v = \begin{pmatrix} 2 \\ -1 \\0 \end{pmatrix}$ supplies that $Av = 0v = 0$ meaning that $0$ is an eigenvalue of $A$ with an eigenvector $v$.


Hint: If a matrix $A$ has column vectors $a_1,..,a_n$, then
$$A\cdot e_i=a_i$$
where $e_1=\pmatrix{1\\0\\0\\ \vdots}$, $e_2=\pmatrix{0\\1\\0\\ \vdots}$, $e_3=\pmatrix{0\\0\\1\\ \vdots}$...
