I would appreciate if someone could help me with the problem in matrix analysis I have been reading for many days. I truly appreciate any help. The problem is this,
Let $A \in M_n$ be Hermitian, let $a_k = \mbox{det} A_k$ be the leading principal minor of $A$ of size $k$ and suppose that all $a_k \neq 0$. Show that the number of negative eigenvalues of $A$ is equal to the number of sign changes in the sequence $+1, a_1, a_2, ..., a_n$. What happens if some $a_i$ is $0$?
For me to solve on this, I use induction, I started with $k = 1$, So $A_k = [a_{11}]$. Thenn $\mbox{det} A_1 = a_1 = a_{11}$ and $\lambda(A_1) = a_11$. Then I took cases if $a_1 > 0$ and $a<0$ and arrived with the conclusion that such claim holds for $k=1$.
I wanted to show for $k = 2$ before going to $k = n$. However, I find so hard to show for the case $k = 2$.
I hope someone could help me with this, I will surely appreciate any help.