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
Goal. (a) Left and right eigenvectors for distinct eigenvalues are orthogonal. (b) A left eigenvector orthogonal to a right eigenvector forces algebraic multiplicity \(> 1\).
1. (a) \(y_i^* x_j = 0\) for \(\lambda_i \neq \lambda_j\).
1.1. \(y_i^* A = \lambda_i y_i^*\) and \(A x_j = \lambda_j x_j\).
\(y_i\) is a left eigenvector and \(x_j\) a right eigenvector.
1.2. \(\lambda_i y_i^* x_j = y_i^* A x_j = \lambda_j y_i^* x_j\).
\(y_i^* A x_j = (y_i^* A) x_j = \lambda_i y_i^* x_j\), and also \(y_i^* A x_j = y_i^* (A x_j) = \lambda_j y_i^* x_j\).
1.3. Hence \((\lambda_i - \lambda_j) y_i^* x_j = 0\), so \(y_i^* x_j = 0\).
\(\lambda_i \neq \lambda_j\).
2. (b) If \(y^* x = 0\) for a left eigenvector \(y\) of \(\lambda\), then \(\operatorname{am}(\lambda) > 1\).
2.1. Suppose \(\operatorname{am}(\lambda) = 1\).
assume for contradiction.
2.2. Then the generalized eigenspace for \(\lambda\) is one-dimensional, spanned by \(x\).
algebraic multiplicity \(1\) means the generalized eigenspace has dimension \(1\).
2.3. The left generalized eigenspace for \(\lambda\) is also one-dimensional, spanned by \(y\).
the left and right generalized eigenspaces for the same eigenvalue have the same dimension.
2.4. The pairing between the left and right generalized eigenspaces for \(\lambda\) is nondegenerate.
the left and right generalized eigenspaces for \(\lambda\) are dual to each other under the pairing \((y, x) \mapsto y^* x\), and this pairing is nondegenerate.
2.5. Hence \(y^* x \neq 0\), contradicting \(y^* x = 0\).
a nondegenerate pairing on a one-dimensional space cannot vanish on the nonzero pair \((y, x)\).
3. Q.E.D.
1 proves (a); 2 proves (b).