Let $A=(a_{ij})$ be a real matrix. Consider the $(n+1)\times (n+1)$ bordered matrix $\tilde A= (a'_{ij})$ where $\tilde a_{ij} = a_{ij}$ for $1\le i,j\le n$, $a'_{ij} = 1- \delta_{ij}$ if $\max(i,j) = n+1$. For instance, if $(a_{ij})$ is $3\times 3$, then the bordered matrix $\tilde A$ is
$$\tilde A=\left(\begin{matrix} a_{11}& a_{12} & a_{13}& 1 \\ a_{21}&a_{22}&a_{23}&1 \\a_{31}&a_{32}& a_{33}& 1\\1&1&1&0\end{matrix} \right)$$
If $A$ is symmetric, then so is $\tilde A$.
Consider the function from real symmetric matrices to real numbers $$\Delta\colon A\mapsto \det \tilde A$$
Note that $\Delta$ is a homogeneous polynomial of degree $n-1$ in the entries of $A$.
I would like to show that if $A$ is a real symmetric matrix -with $\Delta(A)=0$, then for every $\epsilon > 0$ there exists a symmetric matrix $A'$, $\|A'- A\|< \epsilon$, and $\Delta (A')>0$.
Notes:
This is related to my previous question. There, the problem was: $\det A= 0$, find $A'$ close to $A$, with $\det A'>0$. @Martin R: showed us how to find a ray $A + \epsilon L$, $\epsilon>0$ on which the function $\det$ is $>0$. Moreover, if $A$ is symmetric, then $L$ can be chosen to be symmetric too.
Why I am interested in this problem: Consider $A$ a symmetric matrix that is positive definite on the subspace $\sum x_i = 0$. Then there exist $\lambda> 0$ such that the matrix $A + \lambda J$ is positive definite ( $J$ the matrix with all $1$ -- see this question). Consider a principal minor of the matrix $A + \lambda J$. It will be a polynomial of degree $1$ ( since $J$ is or rank $1$) in $\lambda$. The leading term must be $\ge 0$. Now the leading term is exactly $- \times$ one of the determinant in the problem. Moreover, this inequality is true for all $A'$ symmetric, close to $A$ ( since $A_{|V}$ positive definite then same is true for $A'$ close to $A$). Therefore ( based on our still unproved result), the inequalities must be strict.
Thank you for your interest! Any feedback would be appreciated!
