If we have a matrix $$\mathbb{A}= \begin{bmatrix} \mathbb{B} & \mathbb{a}\\ \mathbb{a}^T & 0 \end{bmatrix} $$ where $\mathbb{B}=\mathbb{C}+\sigma \mathbb{aa}^T$ is a $n\times n$ positive semidefinite matrix and $\mathbb{a}$ is a column vector with $n$ elements and $\mathbb{C}$ is also a positive semidefinite matrix. In this case is it always true that $\mathbb{A}$ has exactly one negative eigenvalue? If it is true then how to prove it?
My try:
I know there is theorem (and I do not know how to prove this theorem either and its name either, please provide the name of this theorem) which says that, with the construction of $\mathbb{A}$ provided above, we always have $$\lambda_n(\mathbb{A})\geq \lambda_n(\mathbb{B})\geq 0.$$ But even with this theorem we can say that $n$ eigen values are non-negative but it does not say anything about the $n+1$th eigenvalue. So, I do not know how to use this theorem to prove that $\mathbb{A}$ has exactly one negative eigenvalue. Any help in clarification will be much appreciated. Thanks in advance.
You miss an important assumption that $a$ is nonzero. The form of $B$ does not matter. Assuming that it’s positive semidefinite is enough.
So you know that there’s at most one negative eigenvalue (thanks to the Cauchy interlacing theorem). The whole matrix $A$ cannot be positive semidefinite since otherwise the zero on the diagonal would imply $a=0$ (zero on the diagonal of a semidefinite matrix implies the whole corresponding row/column is zero). Hence it has at least one negative eigenvalue.