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
Best Answer
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.