$A$ and $A'$ have the same eigenvectors: Flaw in the proof?

Question

Consider the equation

$$Ax=\lambda x$$ where $x$ is non zero. Then, we have: $$x'Ax=\lambda x'x$$

$$(x'Ax)'=x'A'x=(\lambda x'x)'=\lambda x'x$$ $$x'(A'x-\lambda x)=0$$ $$A'x=\lambda x$$ (since $x'$ is not zero).

Hence, this proves that A and A' have the same eigenvalues and the same eigenvectors as well.

However, my professor claims that in general, A and A' do not have the same eigenvectors.

What exactly then, is the flaw in my proof?

Answer 1

You cannot deduce from the fact that $x'$ is not $0$ together with $x'(A'x-\lambda x)=0$ that $A'x-\lambda x=0$. For instance, if $A$ is a $2\times2$ matrix, if $x'=(1,0)$ and if $A'x-\lambda x=(0,1)$, then $x'(A'x-\lambda x)=0$, but $A'x-\lambda x\ne0$.