Interlacing of eigenvalues for Hermitian matrices This is a problem from Matrix Analysis by Horn and Johnson.
Let $A \in M_n$ be Hermitian, let $a_k$$=$det$A$[{$1$, $\dots$,$k$}] be the leading principal minor of $A$ of size $k$, $k = 1, \dots, n$, and suppose that all $a_k \neq 0$. Show that the number of negative eigenvalues of $A$ is equal to the number of sign changes in the sequence $+1, a_1, a_2, \dots, a_n$. Explain why $A$ is positive definite if and only if every principal minor of $A$ is positive. What happens if some $a_i =0?$
The hint on the book tells me to use the interlacing theorem, which is the following. 
Let $B \in M_n$ be Hermitian, let $y\in \mathbb C^n$ and $a \in \mathbb R$ be given, and let $A$ $=$ $$ \begin{pmatrix}
        B & y \\
        y* & a \\
        \end{pmatrix}  \in M_{n+1}
$$ 
Then, $\lambda_1(A) \le \lambda_1(B) \le \lambda_2(A) \le \cdots \le \lambda_n(A) \le \lambda_n(B) \le \lambda_{n+1}(A)$.
However, I have not been able to solve this so far and I would greatly appreciate any solution, hint, or suggestions.
 A: See the argument used by Algebraic Pavel here. 
Since $a_{n-1}=\det(B)\neq 0$ then $B^{-1}$ exists. Consider$XAX^*=
\begin{bmatrix}B_{n-1\times n-1}&0_{n-1\times 1}\\0_{1\times n-1}&a-y^*B^{-1}y\end{bmatrix}$, where
$X:=\begin{bmatrix}I_{n-1\times n-1} & 0_{n-1\times 1} \\ -y^*B^{-1} & 1\end{bmatrix}.$
Notice that $\det(X)=1$. 
Therefore,
$a_n=\det(A)=\det(XAX^*)=det(B)(a-y^*B^{-1}y)=a_{n-1}(a-y^*B^{-1}y)$.
Now, if there is a sign change between $a_{n-1}$ and $a_n$ then $a-y^*B^{-1}y<0$.
Then the number of negative eigenvalues of $XAX^*$ is the number of negative eigenvalues of $B$ plus 1. Now, the signatures of $XAX^*$ and $A$ are the same, thus the number of negative eigenvalues of $A$ is the number of negative eigenvalues of $B$ plus 1.
If the signs of $a_{n-1}$ and $a_n$ are equal then $a-y^*B^{-1}y>0$ and the number of negative eigenvalues of $XAX^*$ is the number of negative eigenvalues of $B$.  Since, the signatures of $XAX^*$ and $A$ are the same then the number of negative eigenvalues of $A$ is the number of negative eigenvalues of $B$.
Now, use induction on $B$ to obtain the result.
