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 $v$.


The answer for (I) looks good.
Good for (II) might be to note that
$$
A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}2\\4\\-6\end{bmatrix}
$$
and
$$
A\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}
$$
which gives the first two columns of $A$:
$$
A=\begin{bmatrix}1&2&x\\2&4&y\\-3&-6&z\end{bmatrix}
$$
The trace is the sum of the eigenvalues. Thus, if the sum of the eigenvalues is $0$, we need $z=-5$.  $x$ and $y$ are arbitrary.
