Can we, without further information, find the determinant $$ \det\begin{pmatrix} A & \vec{b} \\ \vec{c}^T & 0 \end{pmatrix} $$ where $A\in\mathbb{C}^{N \times N}; \ \vec{b}, \vec{c} \in\mathbb{C}^N$ and we know that
- $A$ is invertible,
- $\vec{b}, \vec{c}^T$ are right and left eigenvectors respectively with the same non-zero eigenvalue $\lambda$. ($\vec{b}^TA=\lambda\vec{b}^T$, $A\vec{c}=\lambda\vec{c}$),
- $\vec{b}$ and $\vec{c}$ are orthogonal ($\vec{c}^T\vec{b}=0$).
I used the matrix determinant lemma to get $$ \det\begin{pmatrix} A & \vec{b} \\ \vec{c}^T & 0 \end{pmatrix} = 0\det(A) - \vec{c}^T\text{adj}(A)\vec{b} = -\det(A) \ \vec{c}^T A^{-1}\vec{b}. $$ But it feels like there should be some nice result, like that it is $0$ always.