Consider \(\lambda_i, \lambda_j \in \operatorname{eig}(A)\) such that \(\lambda_i \neq \lambda_j\). Let \((x_i, y_i)\) and \((x_j, y_j)\) denote the right and left eigenvectors of \(A\) associated with \(\lambda_i\) and \(\lambda_j\). Show that \(y_i^* x_j = 0\).
Let \(x\) denote an eigenvector of \(A\) associated with an eigenvalue \(\lambda\). Prove that if \(\lambda\) has a left-eigenvector \(y\) such that \(y^* x = 0\), then \(\operatorname{am}(\lambda) > 1\).
Solution
1. If \(\lambda_i\ne\lambda_j\), then \(y_i^*x_j=0\).
Because \(x_j\) is a right eigenvector and \(y_i\) is a left eigenvector, \[ Ax_j=\lambda_jx_j, \qquad y_i^*A=\lambda_i y_i^*. \] Hence \[ \lambda_i y_i^*x_j =y_i^*Ax_j =\lambda_j y_i^*x_j. \] Since \(\lambda_i\ne\lambda_j\), it follows that \(y_i^*x_j=0\).
2. For part (b), suppose toward a contradiction that \(\operatorname{am}(\lambda)=1\).
We will show that the assumed orthogonality \(y^*x=0\) then forces a generalized eigenvector of rank \(2\), contradicting algebraic multiplicity \(1\).
3. Under the assumption \(\operatorname{am}(\lambda)=1\), both the right and left eigenspaces for \(\lambda\) are one-dimensional.
The geometric multiplicity is at least \(1\) and at most the algebraic multiplicity, so \[ \dim\ker(A-\lambda I)=1. \] Also \[ \operatorname{rank}(A-\lambda I) =\operatorname{rank}(A^*-\overline\lambda I), \] so the two kernels have the same dimension. Thus \[ \dim\ker(A^*-\overline\lambda I)=1. \] This latter kernel is the left eigenspace, so the given nonzero left eigenvector \(y\) spans it.
4. The condition \(y^*x=0\) implies\[ x\in\operatorname{im}(A-\lambda I). \]
For every linear map \(B\) on a finite-dimensional inner-product space, \[ \operatorname{im}B=(\ker B^*)^\perp. \] Take \(B=A-\lambda I\). By 3, \[ \ker B^*=\operatorname{span}\{y\}. \] The hypothesis \(y^*x=0\) says exactly that \(x\perp\ker B^*\), hence \(x\in\operatorname{im}B\).
5. There exists \(z\) such that\[ (A-\lambda I)z=x\ne0, \qquad (A-\lambda I)x=0. \]
The first equality follows from 4, and the second from the fact that \(x\) is a right eigenvector for \(\lambda\).
6. The relations in 5 imply that the algebraic multiplicity of \(\lambda\) is at least \(2\).
The vectors \(x\) and \(z\) are linearly independent: if \(z=cx\), then \((A-\lambda I)z=0\), contradicting \((A-\lambda I)z=x\ne0\). Their span is invariant under \(A\), and in the ordered basis \((x,z)\) the restriction of \(A\) has matrix \[ \begin{pmatrix} \lambda&1\\ 0&\lambda \end{pmatrix}. \] Hence \((t-\lambda)^2\) divides the characteristic polynomial of \(A\), so \(\operatorname{am}(\lambda)\ge2\).
7. Therefore \(\operatorname{am}(\lambda)>1\).
This contradicts the assumption in 2 that \(\operatorname{am}(\lambda)=1\).