Left and right eigenvectors cannot be orthogonal 
Let $x, y$ be a right and a left eigenvector corresponding to the same simple eigenvalue (algebraic multiplicity is $1$) of a matrix.
  Show that $x, y$ cannot be orthogonal.

In my opinion, if the eigenvalue with algebraic multiplicity is $1$, that means the power of $(A-\lambda I)$ must be $1$. Does it mean that the eigenvalues must be different? If the eigenvalues are different, then $x,y$ should be orthogonal. So, how to prove $x,y$ cannot be orthogonal?
Thanks a lot.
 A: 
the eigenvalue with algebraic multiplicity is 1, that means the power of $\boldsymbol{A}-\lambda\mathbf{I}$ must be 1

No. That means that the nullity of $\boldsymbol{A}-\lambda\mathbf{I}$ must be $1$, and consequently the index must also be $1$. In brief, $\mathrm{alg}(\lambda)=1$ means $\dim\ker(\boldsymbol{A}-\lambda\mathbf{I})^{n}=1$. For any eigenvalue, we have $\mathrm{geom}(\lambda)\geq1$, that is, $\dim\ker(\boldsymbol{A}-\lambda\mathbf{I})\geq1$. Together, all this implies $$\dim\ker(\boldsymbol{A}-\lambda\mathbf{I})^{n}=\dim\ker(\boldsymbol{A}-\lambda\mathbf{I})^{k}=\dim\ker(\boldsymbol{A}-\lambda\mathbf{I})=1,\quad \forall k\geq1.$$
The least $k$ for which $\ker\boldsymbol{M}^{k+1}=\ker\boldsymbol{M}^{k}$ is called the index of $\boldsymbol{M}$. Equivalent conditions are  $\ker\boldsymbol{M}^{k}\oplus\mathrm{Im}\boldsymbol{M}^{k}$ or, by the rank-nullity theorem, $\mathrm{Im}\boldsymbol{M}^{k+1}=\mathrm{Im}\boldsymbol{M}^{k}$. This is something we will use later.
Now, back to the main question. We have $$\ker(\boldsymbol{A}-\lambda\mathbf{I})=\mathrm{span}\{\boldsymbol{x}\}, \quad \ker(\boldsymbol{A}-\lambda\mathbf{I})^{*}=\mathrm{span}\{\boldsymbol{y}\},\quad \boldsymbol{x},\boldsymbol{y}\neq\boldsymbol{0}.$$

So, how to prove $\boldsymbol{x},\boldsymbol{y}$ cannot be orthogonal?

To show that $\boldsymbol{y}^{*}\boldsymbol{x}\neq0$, suppose the contrary, $\boldsymbol{y}^{*}\boldsymbol{x}=0$, which implies $$\boldsymbol{x}\in\mathrm{span}\{\boldsymbol{y}\}^{\perp}=\ker(\boldsymbol{A}-\lambda\mathbf{I})^{*\perp}=\mathrm{Im}(\boldsymbol{A}-\lambda\mathbf{I}).$$
The existence of $\boldsymbol{x}\neq\boldsymbol{0}$, such that $\boldsymbol{x}\in\mathrm{Im}(\boldsymbol{A}-\lambda\mathbf{I})\cap\ker(\boldsymbol{A}-\lambda\mathbf{I})$, requires that $\mathrm{Im}(\boldsymbol{A}-\lambda\mathbf{I})^{2}\subsetneq\mathrm{Im}(\boldsymbol{A}-\lambda\mathbf{I})$, which contradicts that the index of $\boldsymbol{A}-\lambda\mathbf{I}$ is $1$. Therefore, the only possibility is $\boldsymbol{y}^{*}\boldsymbol{x}\neq0$.
