Question
(This is an updated version of a question I had asked before which remains unsolved). Say for any square matrix A with diagonal elements $a_{ii} =1$ and off-diagonal elements $a_{ij} \in (-1,0]$, for $j\neq i$. Matrix $B = A^TA$. So Matrix $B$ is symmetric. Matrix $C$ is defined as $2diag(B)+B$ which means that the diagonal elements of $B$ are multiplied by 3 while the off-diagonal by 1 (i.e. $c_{ii}=3b_{ii}, c_{ij}=b_{ij} \text{ for } j\neq i$). How do I show that $C$ is invertible?
I managed to solve the 2 by 2 case by comparing common terms in the determinant. But even a 3 by 3 matrix would make the comparison infeasible. Grateful for any comments and suggestions!
As you noted, $B$ is symmetric. Furthermore, it is positive semidefinite since $$x^TBx = x^TA^TAx = \|Ax\|_2^2\geq 0.$$ Define $D = \mathrm{diag}(B)$. Then for the matrix $C$, we have $$x^TCx = x^TBx + 2x^TDx = \|Ax\|_2^2 + 2\|D^{1/2}x\|_2^2\geq 2\min\{\sqrt{d_{ii}}\}\|x\|_2^2\geq \|x\|_2^2.$$ The last inequality comes from the fact that $d_{ii} = \|a_i\|_2^2\geq 1$. Furthermore, notice that equality only holds when $x=0$ since $d_{ii} = \|a_i\|_2^2\geq 1>0$. This inequality implies that $C$ has only positive eigenvalues. Indeed, let $v,\lambda$ be an eigenpair of $C$, then $$v^TCv = \lambda\|v\|_2^2>\|v\|_2^2\implies\lambda>0.$$ Here we used strict inequalities since $v\neq 0$ from the definition of an eigenvector.