When does $A[x,y,z]^T=[y,2x,0]^T$ have a nonzero solution? $$A[x,y,z]^T=[y,2x,0]^T$$ has a nonzero solution
if (and only if?)
$$\det(A-B) = 0$$
where
$$B = \begin{bmatrix}
0 & 1 & 0\\ 
2 & 0 & 0\\ 
0 & 0 & 0
\end{bmatrix}$$
Is that right?

Note that $$B\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} = \begin{pmatrix}
0 & 1 & 0\\ 
2 & 0 & 0\\ 
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=\begin{pmatrix}
y\\ 
2x\\ 
0
\end{pmatrix}.$$ Thus
$$A\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} =\begin{pmatrix}
y\\ 
2x\\ 
0
\end{pmatrix}\iff (A-B)\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=\begin{pmatrix}
0\\ 
0\\ 
0
\end{pmatrix}.$$ Now, the homogenous system $$ (A-B)\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=\begin{pmatrix}
0\\ 
0\\ 
0
\end{pmatrix}$$ has a non-trivial solution iff $\det(A-B)=0.$
