Let $ A \in \mathbb R^n \times \mathbb R^n $ and $ \lambda \in \mathbb C $ show that $ \lambda* $ is also an Eigenvalue. Is my solution acceptable?

Question

I solved this question: Let $ A \in \mathbb R^n \times \mathbb R^n $ and $ \lambda \in \mathbb C $ show that $ \lambda* $ is also an Eigenvalue.

so the official solution to this was a bit complicated and I had a different one.

My Solution was very simple, I wonder if it's valid.

Since the determinant is equal to the product of the eigenvalues and the determinant is real the conjugate must also be an eigenvalue to cancel the imaginary part. Because otherwise we'd have a complex determinant in Real Space. Would this be valid? Thank you!

Answer 1

No, it is not valid. If the determinant is $0$, then the eigenvalues could be, for instance, $0$, $1$, and $i$.

And even if the determinant is not $0$, your argument doesn't work. What prevents the eigenvalues from being $1+i$ (twice) and $2i$ (once)?