Do eigenvectors of a matrix and its transpose, for different eigenvalues, annihilate each other?

Question

The typical proof that I have seen goes like this,

Take $x$ to be an eigenvector of $A$ with eigenvalue $a$, and take $y$ to be an eigenvector of $A^T$ with eigenvalue $b$. Assume that $a \ne b$.

Then, by definition of the transpose (or adjoint), we have that

$$<Ax, y> = <x, A^Ty>$$ $$\iff<ax, y> = <x, by>$$ $$= a<x,y> = \bar{b}<x,y>$$ $$\implies <x,y> = 0$$ as $a \ne b$.

However, as you see: I've used the fact that the inner product is conjugate linear in the second argument, so the above "proof" doesn't seem to work, since $a$ could be equal to $\bar{b}$, and thus $<x,y> = 0$ is not necessarily true.

Where is my mistake in reasoning?

(For instance, the above proof is used in Peter Lax's Linear Algebra book.)

Answer 1

$A=\pmatrix{i & 0\cr 0&-i}$ in the orthogonal basis its transpose $\pmatrix{-i & 0\cr 0& i}$ but $e_1$ is not orthogonal to $e_1$.

You use the fact that $\langle A(x),y\rangle=\langle x, A^t(y)\rangle$, this is true for the conjugate transpose not for the transpose. Since $\langle A(e_1),e_1\rangle=i$ and $\langle e_1,A(e_1)\rangle=-i$.

Answer 2

They only annihilate when $a\ne b.$ So, the assumption $a\ne b$ is fine.

In the example above, if $a=b$ then $<x,y> \ne 0$

Here is annother thought...

$A = PDP^{-1}$

$A^T = {P^{-1}}^TDP^{T}$

The eigenvectors or $A^T$ are the columns of ${P^{-1}}^T$

Or the rows of $P^{-1}$

And when $n\ne m$ the $n^{th}$ row anihilates the $m^{th}$ column.