I am interested to know the eigenvalues of the following block matrix, which I obtained from the Lagrange function (optimization). We have $H_{n\times n}$ here positive definite (i.e., all the eigenvalues of $H$ are positive).
$$\begin{bmatrix} H& C^T\\C&0\end{bmatrix}$$
Can we say that all eigenvalues of this matrix are positive under some suitable restrictions on $C_{m\times n}$?
Also, if this structure of matrix is well-known in matrix theory please let me know or refer.
Thank you very much for your time.
Assuming $H$ is symmetric, this matrix (let's call it $M$) is also symmetric, and so its eigenvalues are positive iff we have $x^TMx > 0$ for all $x \in \mathbb{R}^{n+m}$, $x \neq 0$.
If we now write $x = \pmatrix{x_1 \\ x_2}$, with $x_1 \in \mathbb{R}^n$, $x_2 \in \mathbb{R}^m$, then we get: $$\begin{split} x^TMx & = \pmatrix{x_1^T & x_2^T}\pmatrix{H & C^T\\C & 0} \pmatrix{x_1 \\ x_2}\\ & = \pmatrix{x_1^T & x_2^T}\pmatrix{Hx_1 + C^Tx_2 \\ Cx_1}\\ & = x_1^THx_1 + x_1^TC^Tx_2 + x_2^TCx_1\\ & = x_1^THx_1 + 2 \cdot x_2^TCx_1 \quad\,\,\text{since}\,\, x_1^TC^Tx_2 = \big(x_2^TCx_1\big)^T\end{split}$$ Therefore, if $x_1 = 0$, but for any $x_2$: $$x^TMx = 0$$ Thus $M$ cannot be positive definite no matter what $C$ is, meaning one of the eigenvalues is indeed non-positive like @sss89 said above in the comments.
Moreover, with the right choice of $x_2$ and $C \neq 0$ you might be able to find $x$ such that $x^TMx < 0$, guaranteeing one negative eigenvalue? Though I'm less sure about that one. One concrete example of this is if $m = n$, $C = H$ and $x_2 = -x_1$, then you do get what I just said, because then $x^TMx = -x_1^THx_1 < 0$.