Let $A$ be a non-normal square matrix and suppose $v, \lambda$ is an eigenpair of $A$, that is, $A v = \lambda v$. Consider the two statements
- $|\lambda| = 1$.
- $A^* A v = A A^* v = v$.
Under which conditions do either of these imply the other?
If the matrix $A$ were normal, then these are equivalent. Indeed, consider the projection of $A$ onto the eigenspace spanned by $v$, namely $A|_v = \lambda v v^*$. Then, the following holds:
$$ A A^* v = A^* A v = (\lambda \bar{\lambda}) (v v^*) (v v^*) v = (\lambda \bar{\lambda})v. $$
Thus, the two statements above imply each other. This proof does not work in the non-normal case because the eigenprojector of $A$ onto $v$ is $\lambda v u^*$, where $u$ is the corresponding left eigenvector.
Any help is appreciated.
EDIT: Note statements 1. and 2. are assumed to apply only to the select eigenpair $v, \lambda$. In particular, it may be the case that $A$ has other known eigenvalues that are not unitary. More specifically, statements 1. and 2. are not quantified over each eigenvector, but only to one, namely $v$. I am essentially asking when a non-normal matrix behaves like a unitary matrix when projected on a particular eigenspace.
In general, 2 implies 1, but not the other way around.
Proof: $Av = \lambda v$ means that $v^*A^* = \bar{\lambda} v^*$, so $$v^*v = v^*(A^*Av) = (\bar{\lambda} v^*)(\lambda v) = |\lambda|^2 v^*v$$ Since $v$ is nonzero, $|\lambda|=1$.
Counterexample to 1 implies 2: The matrix $A = \begin{bmatrix}1&1\\0 &1\end{bmatrix}$ has eigenvector $v = \begin{bmatrix}1\\0\end{bmatrix}$ with $\lambda = 1$, but $AA^*v = \begin{bmatrix}2\\1\end{bmatrix}$, $A^*Av = \begin{bmatrix}1\\1\end{bmatrix}$, and $v = \begin{bmatrix}1\\0\end{bmatrix}$ are not equal to each other.