Suppose we consider an equation $A f=\lambda f^*$. Here $A$ is a differential operator with the general form $$A=a\frac{d^2}{dt^2}+b g(t)$$ where $a,b\in\mathbb{C}$ and $g(t)$ is a complex function of real variable $t$. $f$ is complex function $f=f(t)$, and $f^*$ is its complex conjugate. $\lambda$ is a real number.
We have a complex-conjugated equation $A^* f^*=\lambda f$. In total, we have the matrix form $$\pmatrix{\ 0\ \ A^*\\ A\ \ 0}\pmatrix{f\\ f^*}=\lambda\pmatrix{f\\ f^*}.$$
It is very easy to see that we have another equation $$\pmatrix{0\ \ A^*\\ A\ \ 0}\pmatrix{if\\ -if^*}=-\lambda\pmatrix{if\\ -if^*}.$$
Now I will conclude that $$-\lambda^2=\det\pmatrix{0\ \ A^*\\ A\ \ 0}=-\det(A^*A).\tag{1}$$
All these look fine to me up to now. But I tried another way in the following and found a disagreement. Writing $A=U+iV$, $f=x+iy$, then we have the following matrix equation $$\pmatrix{U\ \ -V\\ -V\ \ -U}\pmatrix{x\\ y}=\lambda\pmatrix{x\\ y}.$$ This equation is also associated with another equation $$\pmatrix{U\ \ -V\\ -V\ \ -U}\pmatrix{-y\\ x}=-\lambda\pmatrix{-y\\ x}.$$ And I conclude $$-\lambda^2=\det\pmatrix{U\ \ -V\\ -V\ \ -U}=-\det (U^2+V^2).$$ However $A^*A=U^2+V^2+i[U,V]$. If both methods are correct then there must be $$\det (U^2+V^2+i[U,V])=\det(U^2+V^2).\tag{2}$$ Is this true? Or there is something wrong in one of these two methods or in both?
For Eq.(2), one possibility that it may be true is that we know $$\det(AA^*)=\det(A^*A)$$ which gives $$\det(U^2+V^2+i[V,U])=\det(U^2+V^2+i[U,V]).$$ Somehow the part $i[V,U]$ or $i[U,V]$ does not contribute in the determinant?