Positive definiteness of matrix in special case

Question

Let $A \in \mathbb{R}^{n \times n}$, $C \in \mathbb{R}^{m \times m}$ be symmetric and let $B \in \mathbb{R}^{m \times n}$ such that $B^T \mathbf{x} \neq \mathbf{0}$ for all $\mathbf{x} \in \mathbb{R}^m \setminus \{\mathbf{0}\}$. We consider the matrix $$K=\begin{bmatrix} A & B^T\\ B & C \end{bmatrix}.$$
I have shown that $K=LDL^T$, where $$L=\begin{bmatrix} I_n\\ Q & I_m \end{bmatrix},D=\begin{bmatrix} H\\ & S \end{bmatrix},$$ where $H=A$, $Q=BA^{-1}$ and $S=C-BA^{-1}B^T$.
I have also shown that $K$ is positive definite if and only if both $H$ and $S$ are positive definite.
The question is: in the case when $C=O$, is $K$ positive definite? Is it negative definite? Is it indefinite?
So if $C=O$, we have $S=-BA^{-1}B^T$ and $$K=\begin{bmatrix} A & B^T\\ B & O \end{bmatrix}.$$
I think it is indefinite, but I am not sure how to proceed.

Answer 1

As $B\ne0$, we have $b_{ij}\ne0$ for some $i$ and $j$. Therefore $K$ contains a principal submatrix $\pmatrix{a_{jj}&b_{ij}\\ b_{ij}&0}$ that is indefinite because the product of its two eigenvalues (i.e. its determinant) is negative. Hence $K$ is indefinite.

Answer 2

The matrix $K$ is of size $(m+n)\times (m+n)$ with lower right $m\times m$ part being the zero matrix (C in your notation).

Take $v= e_{n+i}$ for $i=1,2,\ldots m$ (standard basis vectors)

Then simple matrix multiplication yields $Kv=0$. So $v^tKv=0$, violating positive definiteness.

Answer 3

Seek the help of Sylvester. The matrix $K$ is congruent to $D$ and hence they have the same inertia (number of positive, negative, and zero eigenvalues). This has two consequences:

1. $K$ is positive definite if and only if $D$ is which is positive definite if and only if $H$ and $S$ are.

2. If $C$ is zero, $S$ is negative definite and hence $K$ is indefinite. In addition, $K$ has $n$ positive and $m$ negative eigenvalues.