I've been working on the following question with little succes for the past few days and could use some help of a kind stranger

For a symmetric matrix $X$ with the $k$-th principal minor of $X$ noted as $X_k$ (so consisting of the first $k$ rows and columns of $X$). Suppose $X$ is such that $\det(X_k)>0 \forall k\in [n-1]$, but $\det(X)<0$. Define $$a_i = (-1)^i M_{(i,n)}$$ where $M_{(i,n)} = \det(X_{(i,n)})$ where $X_{(i,n)}$ is $X$ without the $i$-th row and $n$-th column. I wnat to show $$a^Ta = \det(X)\det(X_{n-1})$$

I've already shown that $$\det(X)\det(X_{n-1}) = a_n\big(\sum_{i=1}^{n}aix_{n,i}\big)$$ (with $x_{ij}$ just the $i,j$-th entry of $X$) and that $X_k$ is positive definite for all $k=1,\dots n-1$ because of Sylvesters criterion.

I've been trying to write out $a^TXa$ for hours but i havent come further than $$a^TXa > 2\det(X)\det(X_{n-1}) - a_n^2x_{n,n}$$ but thats a little useless since you don't want to strictly bound it... I was wondering what I could say about $x_{n,n}$ knowing all that stuff about the determinants and stuff.

What do I do?



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