Eigenvector and adjoint eigenvector not orthogonal Let $A\in M_n (\mathbb{C})$ for which
$$
Aq = \lambda q,
$$
where $q$ is its eigenvector corresponding to a simple eigenvalue $\lambda$.
Consider the adjoint eigenvector $p$:
$$
A^* p = \bar{\lambda} p.
$$
How does one prove that $(p, q)\neq 0$ where $(\cdot,\cdot)$ stands for the standard scalar product on $\mathbb{C}^{n}$? In other words, $p$ and $q$ are not orthogonal?
I came across this assertion in Elements of Applied Bifurcation Theory [Yuri A. Kuznetsov] on page 92.
 A: Let $A \in M_n(\mathbb{C})$ be a matrix such that $0$ is a simple eigenvalue of $A$ (that is, the algebraic multiplicity of $0$ is one). Let $q \neq 0$ such that $Aq = 0$ and $p \neq 0$ such that $A^{*}p = 0$. Assume that $p \perp q$. Then 
$$q \in \operatorname{span} \{ p \}^{\perp} = \ker(A^{*})^{\perp} = \operatorname{Im}(A) $$
so write $q = Av$ for some $v \neq 0$. Then $A^2v = Aq = 0$ and $q,v$ are linearly independent so $\dim \ker(A^2) \geq 2$ which implies that the algebraic multiplicity of $0$ is $\geq 2$, a contradiction.
For your situation, apply the above to $A - \lambda I$ and conclude that $p,q$ can't be perpendicular.
A: Assume for the moment that $(p,q)=0$. There exists a unitary matrix $U$ whose first column is $q$ and second column is $p$. Then
$$
A_U := U^* A U = 
\begin{pmatrix}
\lambda & + & + \\
0 & \lambda & 0 \\
0 & + & B
\end{pmatrix},
$$
where $+$ denotes arbitrary matrix elements, and $B$ is a matrix of dimension $(n-2) \times (n-2)$. The eigenvalues of $A_U$ are the same as those of $A$. However, note that the characteristic polynomial of $A_U$ is 
$$
p_{A_U}(t) = (\lambda - t)^2p_B(t).
$$
Hence, the algebraic multiplicity of $\lambda$ is at least two and $\lambda$ is not simple.
