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

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?

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.
• Nice proof! Note that statement is trivially true also for $\sigma=0$. Commented Jan 23, 2023 at 21:47