Existence of a non-singular $n-2$ principle minor Suppose $A$ is a $n\times n$ real symmetric matrix with $|A|\ne0$, and all the principal minors of order $n-1$ have determinant zero.
Prove: there is a principal minor of order $n-2$ which is non-zero, and if some principal minor of order $n-2$ is non-zero, then it must have opposite sign with $|A|$
A few words about the background: this is one problem in our final exam on linear algebra, I'm the assistant of the course teacher and I found almost no student can give a  precise answer, though some of them did get very near to the solution. Also the standard solution given by us is quite involved, so I posted it here and hope someone might given a short proof.
The statement of the problem is very short, so one may hope it does have a short proof too.    
 A: Geometrically, it's equivalent to:

Suppose $E=\{e_1,\dotsc,e_n\}$ is a basis for a vector space $V$ over $\mathbb R$, and $B$ is a symmetric bilinear form on $V$. If $B$ is nondegenerate but degenerates on every subspace spanned by $n-1$ vectors of $E$, then
  
  
*
  
*$B$ should be nondegenerate on a subspace $W$ spanned by $n-2$ vectors of $E$, and
  
*for any such $W$, let $X,Y$ be the matrices for $B$ and $B\rvert_W$ for any basis of $V$ and $W$ respectively, then $\det X\det Y<0$.
  

Remark on 2: Generally, if $k$ is a field and $k^*$ is the multiplication group of invertible elements. Denote $k^{*2}=\{x^2\colon x\in k^*\}$ as a subgroup of $k^*$, then the discriminant of a nondegenerate bilinear form $B$ is defined as the image of $\det M$ in the quotient group $k^*/k^{*2}$, where $M$ is the matrix of $B$ with respect to any basis. It's routine to check that the discriminant is well-defined, i.e., independent of choice of the matrix (cf. Morandi's Field and Galois Theory, section 12, page 120). Thus it's free for us to choose a specific basis in 2.
To start with, we observe that in any quadratic space $V$,

If $W\subseteq V$ with codimension $1$, then $V^\perp\subseteq W$ or $W^\perp\subseteq V^\perp$ (and hence $W^\perp=V^\perp$).

Indeed, if $V^\perp\not\subseteq W$, we choose $v\in V^\perp\setminus W$, and then $V=\langle v\rangle\oplus W$. For each $w\in W^\perp$, $B(v,w)=0\implies w\in V^\perp$.
Let's proceed with 1. Suppose not, i.e. $B$ degenerates on each subspace spanned by $n-2$ vectors of $E$, we'll show that $B$ degenerates on the whole space $V$. Let's temporarily consider the subspace $V_1=\langle e_1,\dotsc,e_{n-1}\rangle$, and $\hat B=B\rvert_{V_1}$. We can suppose, without loss of generality, that $V_1^{\perp\hat B}\not\subseteq W$ for $W=\langle e_1,\dotsc,e_{n-2}\rangle$, since $\hat B$ degenerates and the intersection of all subspaces spanned by $n-2$ vectors of $\{e_1,\dotsc,e_{n-1}\}$ is $\{0\}$, then by the preceding observation, $W^{\perp\hat B}=V_1^{\perp\hat B}\neq\{0\}$. Let's consider $\langle e_1,\dotsc,e_{n-2},e_n\rangle$. Again by the same observation, we can choose $0\neq w\in W$ such that $B(w,e_k)=0$ for $k=1,\dotsc,n-2,n$, hence $w\in W^{\perp\hat B}$ and $B(w,e_{n-1})=0$, therefore $B$ degenerates.
Now consider 2. WLOG, suppose $B$ is nondegenerate on $W=\langle e_1,\dotsc,e_{n-2}\rangle$. We choose an orthogonal basis $\{f_1,\dotsc,f_{n-2}\}$ for $W$. Since $B$ degenerates on $\langle e_{n-1}\rangle\oplus W$, we can apply Gram-Schmidt process to obtain $f_{n-1}$ such that $e_{n-1}-f_{n-1}\in W$, $f_{n-1}\in W^\perp$ and $B(f_{n-1},f_{n-1})=0$. Similarly, we can choose $f_n$ such that $e_n-f_n\in W$, $f_n\in W^\perp$ and $B(f_n,f_n)=0$. Now the proposition is obvious with respect to the basis $\{f_1,\dotsc,f_n\}$.
Further remark:
In the preceding proof, we can see that the discriminant $\operatorname{disc}(B\rvert_W)=(-1)\operatorname{disc}(B)$ for arbitrary field $k$.
