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
 A: This is usually proved by using matrix congruence, but yes, you can prove the statement by using the adjugate matrix.
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.
A: 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.
