When does $A[x,y,z]^T=[y,2x,0]^T$ have a nonzero solution?

Question

$$A[x,y,z]^T=[y,2x,0]^T$$ 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?

Answer 1

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 $\det(A-B)=0.$