This may be a basic question, but I'd like someone to double check it.
We are given the matrix $A=\begin{pmatrix} A_1 & C \\ C^T & A_2\end{pmatrix}$ where $A_1$ is a $k$ by $k$ positive definite matrix with entries from $\mathbb R$, $A_2$ is a $l$ by $l$ negative definite with entries from $\mathbb R$, $C$ is some matrix, and $C^T$ is $C$ transposed.
Show that the signature of $A$ is $(k,l,0)$
My proof
We know $A_1$ is positive definite, and so it has $k$ positive eigenvalues (not necesarily different).
$A_2$ is negative definite, and so it has $l$ negative eigenvalues (again, not necessarily different).
The determinant of a block matrix is the product of the determinants of the blocks on the diagonal, so:
$det(A-\alpha I) = det\begin{pmatrix} A_1-\alpha I & C\\ C^T & A_2-\alpha I\end{pmatrix} = det(A_1-\alpha I)det(A_2- \alpha I)$
We know that $det(A_1-\alpha I)det(A_2- \alpha I)=0$ if and only if $det(A_1-\alpha I)=0$ or $det(A_2- \alpha I)=0$.
So, we can infer that since $A_1$ has $k$ positive eigenvalues, and $A_2$ has $l$ negative eigenvalues, $A$ also has $k$ positive and $l$ negative eigenvalues.
since those values are the only values for which $det(A)=0$, it has no eigenvalue which is equal to zero, and so the signature of $A$ is $(k,l,0)$