Eigenvalues of a structured matrix Consider the block matrix \begin{pmatrix} A & b \\ c^T & d \end{pmatrix} where $A \in \mathbb{R}^{n \times n}$ is a diagonal matrix, $b,c \in \mathbb{R}^n$ and $d \in \mathbb{R}$, with $n > 1$.
What do we know about the eigenvalues of this matrix? I am specifically looking for conditions ensuring that they are all in the open left half of the complex plane (i.e. the associated LTI dynamical system is stable).
 A: Case 1) $A$ is invertible:
You can use the determinant for block matrices: see here.
Call your matrix $G$. Consider $G-\lambda\mathbb{1}_{n+1}$ which is of the same form
$$\begin{pmatrix} A-\lambda\mathbb{1}_n & b \\ c^T & d-\lambda \end{pmatrix}$$
Then $\det(G-\lambda\mathbb{1}_{n+1})=\det(A-\lambda\mathbb{1}_n)\det(d-\lambda-c^TA^{-1}b)$.
Now letting $A=\operatorname{diag}(a_1,a_2,\ldots,a_n)$, and noting that inside the second determinant is just a $1$ by $1$ matrix or simply a number, we have:
$\det(G-\lambda\mathbb{1}_{n+1})=(d-\lambda-c^TA^{-1}b)\prod\limits_{i=1}^n (a_i-\lambda)$
So $n$ of the eigenvalues are $\{a_1,\ldots,a_n\}$, and the last is obtained by computing $\lambda=c^TA^{-1}b-d$, and $A^{-1}=\operatorname{diag}(a_1^{-1},\ldots,a_n^{-1})$.
Case 2 $A$ is not invertible:
We can instead use that $\det(G-\lambda\mathbb{1}_{n+1})=(d-\lambda+1)\det(A-\lambda\mathbb{1}_n)-\det(A+bc^T)$.
In that case you've to solve a polynomial. I'm not sure if this will simplify further otherwise.
A: The solution provided is not correct. Indeed, generally, the $a_i$'s are not eigenvalues of the whole matrix. For a counter example (and correct answer) see Robert Israels's answer for this question.
Note: This is written as an answer, because I cannot comment.
