Orthogonality rules for a non-diagonalizable matrix Consider a non-diagonalizable matrix $H$ of size 2x2 with one nonderogatory eigenvalue $\lambda$. Therefore, we have that
$$Hu_0=\lambda u_0\quad \mathrm{and}\quad Hu_1=\lambda u_1+u_0$$ where $u_0$ is a right eigenvector and $u_1$ a Jordan right eigenvector.
Similarly, we have that
$$v_0^*H=\lambda v_0^*\quad \mathrm{and}\quad v_1^*H=\lambda v_1^*+uv_0^*$$
where $v_0^*$ is a left eigenvector and $v_1^*$ a Jordan left eigenvector.
How to prove that $v_1^*u_0\ne 0$?

Original post in the quantum mechanics formalism.
Consider a non-diagonalizable matrix of size 2x2, then (it will have degenerate eigenvalues and only one eigenvector)
So,
$H|\phi>=\lambda|\phi>$
and
$H|\phi^J>=\lambda|\phi^J>+|\phi>$
|> just denotes a vector and <| its' dual (bra-ket notation)
here
$|\phi^J>$ represents the associated jordan vector
consider that for $H^\dagger$
$H^\dagger|\xi>=\lambda^{*}|\xi>$
and
$H|\xi^J>=\lambda^{*}|\xi^J>+|\xi>$
here $\lambda^*$ represents complex conjugate of $\lambda$
How to prove,
$<\xi^J|\phi> \neq 0$
 A: So, we can evaluate
$$v_0^*u_0=v_0^*(H-\lambda I)u_1=0$$
where we have used the fact that $v_0^*(H-\lambda I)=0$.
Now we can evaluate
$$v_1^*u_0=v_1^*(H-\lambda I)u_1=v_0^*u_1=\alpha$$
for some $\alpha\in\mathbb{R}$, where we have used the fact that $u_0=(H-\lambda I)u_1$ and $v_0^*=v_1^*(H-\lambda I)$.
Now we note that $u_0$ and $u_1$ as well as $v_0$ and $v_1$ are linearly independent. To see this, let us assume that $u_1=cu_0$ for some $c\ne 0$ and $u_0\ne0$. Then, we have $(H-\lambda I)u_1=(H-\lambda I)cu_0=0$ from the definition of $u_0$, which leads a contradicion.
Therefore the matrices
$$V = \begin{bmatrix}v_0^*\\v_1^*\end{bmatrix}\qquad \mathrm{and}\quad U=\begin{bmatrix}u_0 & u_1\end{bmatrix}$$
form bases of $\mathbb{R}^2$ and are thus invertible. Therefore, $UV$ must be invertible as well. Evaluating this product yields
$$VU = \begin{bmatrix}v_0^*u_0 & v_0^*u_1\\v_1^*u_0 & v_1^*u_1\end{bmatrix}=\begin{bmatrix}0 & \alpha\\\alpha & v_1^*u_1\end{bmatrix}.$$
The product $UV$ is invertible if and only if $\alpha\ne 0$. This proves the result.
