Proving that two right and left eigenvectors are not orthogonal. Let $A$ be a square non-Hermitian matrix and $c$ be an eigenvalue of $A$ with algebraic multiplicity $1$. Let $Ax = cx$ and $y^{H}A = cy^{H}$ where $y^{H}$ is a conjugate transpose of $y$. Prove that $y^{H}x \neq 0$. 
Please give me a hint. Thanks.
 A: EDIT:@me10240 pointed out a flaw in the porevious proof. So i am giving a new proof.
Let say $y^Hx=0$. Then $x\in \mathcal{R}(A-cI)$ since $y\in \mathcal{N}(A^H-c^*I)$. Then, $\exists u\ne 0$ such that $x=(A-cI)u\Rightarrow (A-cI)^2u=0$ Now, since $c$ has algebraic multiplicity $1$, we have $\dim(\mathcal{N}(A-cI)^2)=\dim(\mathcal{N}(A-cI))=1$. Hence, $u=\alpha x$ for some $\alpha\ne 0\Rightarrow x=(A-cI)\alpha x=0$ which is a contradiction.
A: I suppose you forgot to say that $x,y$ are nonzero, otherwise this is not true of course. On the other hand the hypothesis about being non-Hermitian is a red herring, and the conjugate-transpose in $y^H$ serves no purpose: $y^H$ is just a complicated name for an arbitrary $1\times n$ matrix, a linear form on $V=\Bbb C^n$.
Choose a basis such that the linear form $y^H$ is the final coordinate function, i.e., start with choosing a basis $[e_1,\ldots,e_{n-1}]$ of $W=\ker(y^H)$ and complete with a final vector $e_n$ for which $y^He_n=1$. Now the fact that $$y^HAe_i=cy^He_i=c\delta_{i,n}$$ shows that after change of basis $A$ to this new basis $[e_1,\ldots,e_n]$, the final row of the new matrix $A'$ is $\begin{pmatrix}0&\ldots&0&c\end{pmatrix}$ (since that final row is the result of composing the final coordinate function $y^H$ with $A$, and then evaluating it on all basis vectors $e_1,\ldots,e_n$). Then the characteristic polynomial$~\chi$ of $A'$ (and therefore of $A$) is the product of the characteristic polynomial$~\chi'$ of the top-left $(n-1)\times(n-1)$ submatrix$~B$ of $A'$ (which gives the restriction of $A$ to the subspace$~W$) and a factor $X-c$ (for the bottom right $1\times1$ submatrix of$~A'$). Since $c$ is a simple root of$~\chi$, it is not a root of $\chi'$. This means there is no eigenvector of$~A$ for$~c$ in the subspace$~W$. Since $x$ is an eigenvector of$~A$ for$~c$, one has $x\notin W$ as desired.

I've included quite a bit of detail, but if you are used to less matrix-centred reasoning, you can formulate this argument in a single sentence: by hypothesis $c$ is a simple eigenvalue of the operator that $A$ defines on$~V$, and it is also the eigenvalue of the operator it defines on the $1$-dimensional quotient module $V/W$, but then it cannot also be an eigenvalue of the restriction to$~W$.
