# 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.

• Possible answer at this question. Apr 19, 2020 at 10:33

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 $$\boldsymbol{y}^{*}\boldsymbol{x}\neq0$$.