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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 $(k,l,0)$

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"The determinant of a block matrix is the product of the determinants of the blocks on the diagonal"

this statement is incorrect. Simply take

$det(A) = det\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} = 0 \ne 1*1 $

So, the proof does not work as stated.

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