Sylvester's Criterion for indefinite matrices

Question

My definition of indefinite matrix is a (Hermitian) matrix with both positive and negative eigenvalues (in particular it is invertible). How can I use the leading principal minors to identify an indefinite matrix? For example take the matrix $A=\begin{pmatrix} 0&4\\4&6\end{pmatrix}$, then $\Delta_1=0$ and $\Delta_2=-16$, and the eigenvalues are $8$ and $-2$, so it is indefinite.

Answer 1

For a Hermitian matrix $H$ to meet your definition, it must satisfy all the following 3 conditions. $$\text {(i) $\exists$ negative principal minor}$$ $$\text {(ii)$\exists$ positive principal minor of odd order}$$ $$\text {or $\exists$ negative principal minor of even order}$$ $$\text {(iii)} \det (H) \ne 0 $$

Answer 2

You cannot. E.g. all leading principal minors of $A=\pmatrix{0&0&0\\ 0&0&1\\ 0&1&0}$ are zero, but $A$ is indefinite.

By Sylvester's criterion, a Hermitian matrix $A$ is positive semidefinite (resp. negative semidefinite) if all principal minors (not just the leading ones) of $A$ (resp. $-A$) are nonnegative. Therefore, to check that whether $A$ is indefinite or not, you need to check whether $A$ has two principal minors of different signs.

Remark. In your (very uncommon) definition of indefinite matrix, you require invertibility. So, you must also check that $\det(A)\ne0$.