# Real symmetric matrix of rank $n-1$ has a submatrix of rank $n-1$

Let $$A$$ be a symmetric real matrix of dimension $$n \times n$$ and rank $$n-1$$. Prove that there is a $$k \in \{1,2,...n\}$$ such that on deletion of the $$k$$th row and column the resulting matrix has rank $$n-1$$.

I think we would have to use adjugate of the matrix here since that is the space of all $$(n-1) \times (n-1)$$ submatrices, but I am not very sure of how to proceed

As $$A$$ has rank $$n-1$$, it adjugate matrix has rank one. Since $$A$$ is also symmetric, so must be its adjugate matrix. Therefore $$\operatorname{adj}(A)={\pm vv}^T$$ for some nonzero vector $$v$$. Thus $$\operatorname{adj}(A)$$ has some nonzero diagonal entries. As the diagonal entries of $$\operatorname{adj}(A)$$ are the principal $$(n-1)$$-rowed minors of $$A$$, the result follows.
Here is another (slightly) different way to use the fact that the adjugate is a rank $$1$$ symmetric matrix. If it has rank one and furthermore there does not exist some $$k$$ such that the determinant of the $$(k, k)$$th minor is nonzero, then the diagonal of the adjugate matrix (call it $$A$$) is all zeroes.
We know a rank one matrix is simply a matrix whose rows are all multiples of the same non-trivial vector (with at least one row being a nonzero multiple). Suppose the $$i$$th row of $$A$$ were nonzero. Then there exists some $$j \neq i$$ such that $$A_{ij}$$ (recall our assumption is that the diagonal elements are all zero). As the adjoint is symmetric, $$A_{ji} \neq 0$$. However, the $$j$$th row cannot be a multiple of the $$i$$th row (why?), a contradiction. So there cannot be a symmetric matrix of rank $$1$$ whose diagonal elements are all zero, and hence we conclude that at least one element of $$A$$'s diagonal (say the $$k$$th element) must be nonzero. Removing the $$k$$th row and column, of course, will yield an invertible $$(n - 1) \times (n - 1)$$ matrix.