# Hermitian, hollow matrix and invertibility

I have an $(n+m)$x$(n+m)$ block, symmetric, real matrix with all non-negative entries and all zeroes in the main diagonal, with $m<n$. The first block is a $n$x$n$ invertible matrix, and the other diagonal block is a $m$x$m$ matrix of zeroes. The off-diagonal blocks are a $m$x$n$ matrix of rank $m$ and its transpose. Can I guarantee that the matrix is invertible? Can I do it if I also know that the matrix is negative semidefinite? Thanks!

• Do you mean that the dimensions of the entire matrix are $(n+m)\times(n+m)$? – Greg Martin Oct 15 '16 at 1:05
• Hint: Schur complement – user251257 Oct 15 '16 at 2:06
• Yes, sorry, Greg: dimension is $(n+m)$x$(n+m)$. – Rob Oct 15 '16 at 12:31
• Yes, user251257, my question could be phrased differently as: do we know that the Schur complement of the first matrix (the $n$x$n$ non-singular matrix) is non-singular? – Rob Oct 15 '16 at 12:39
• @Rob use the definitness and rank condition – user251257 Oct 16 '16 at 23:02

So, here is my attempt to an answer. Is it correct? The first $n$x$n$ diagonal block $D$ is negative semidefinite, so its inverse is negative semidefinite as well, and the negative of that inverse is positive semidefinite. Then there exists a permutation matrix $P$ such that $-P'D^{-1}P$ has a unique Cholesky decomposition $R'R$, where $R$ is an upper triangular matrix. Since $D^{-1}$ is invertible, $R$ has all diagonal elements positive. Then, if we let $A$, $A'$ be the skew diagonal elements of the block matrix, $-A'D^{-1}A=-A'PP'D^{-1}PP'A=-BR'RB'$, where $B=A'P$. Note that $B$ has full rank ($m$), since $A$ has. That is, its rows are linearly independent. Therefore, $BR'$ has rank $m$, and so has $BR'RB'$. Therefore, we conclude that the Schur complement of $D$ has full rank, and so the block matrix is indeed, invertible.