Control the summation of a diagonal matrix and another matrix to be full rank

Question

Statement. To ensure the rank of $\operatorname{ddiag}(AQQ^T)-\sigma\Delta=n$, it is sufficient to require $\min_i(\operatorname{diag}(AQQ^T))_i>\sigma\lVert\Delta\rVert$.

Note: $Q\in\mathbb{R}_{n\times 2}$, $\sigma$ is a scalar constant, $\Delta$ is $n\times n$ random matrix. The operator $\operatorname{ddiag}$ sets all off-diagonal entries of a matrix to zero. The operator $\operatorname{diag}$ takes the diagonal entries of a matrix.

I don't know why this holds. And also, could someone point out which theory or which direction this statement belongs to, so that next time I can be more clear about where I can find this kind of techniques. Is it just basic linear algebra or something more advanced?

Answer 1

You must assume $\sigma > 0$. Then it's just the fact that the if $D$ is a diagonal $n \times n$ matrix with minimum diagonal entry $m > \|B\|$, where $B$ is an $n \times n$ matrix, $\|D^{-1} B\| \le \|D^{-1}\| \|B\| < 1$, so $I - D^{-1} B$ is invertible, and then $D - B = D (I - D^{-1} B)$ is invertible.