I stumbled across this equation in a paper, which may seem obvious, but I'm wondering if someone can explain why this is true?
By definition of an operator norm,
$$\left[(D^*D)^{-1} - I_d\right]_{\operatorname{op}} = \sup_{\Vert y \Vert = 1} \Vert(D^*D)^{-1}y - y \Vert = \sup_{\Vert y \Vert = 1} \left\vert \Vert(D^*D)^{-1}y\Vert - \Vert y \Vert \right\vert$$
Here $D$ is a fourier ensemble and $I_d$ is the identity matrix. But why does the last equality hold?
Let $A = (D^* D)^{-1}$ and note that $A$ is Hermitian and positive definite, hence unitarily diagonalizable. That is, there is some unitary $U$ such that $UAU^* = \Lambda$, where $\Lambda$ is a diagonal matrix of positive reals. Since $U$ is unitary (and I am presuming the Euclidean norm here), we have $\|A-I\|= \|\Lambda-I\|$, and so $\|A-I\| = \max_k |\lambda_k-1|$, where $\lambda_k$ are the diagonal elements of $A$. In particular, $\|A-I\| = |\lambda^*-1|$ for some diagonal element $\lambda^*$.
If $\|x\| \le 1$ we have $\|A-I\| \ge \|Ax-x\| \ge |\|Ax\|-\|x\||$ and so $\|A-I\| \ge \sup_{\|x\| = 1} |\|Ax\|-\|x\||$.
If we choose $x$ to be a unit eigenvector corresponding to $\lambda^*$, we have $\|Ax\| = \lambda^*$, hence we have equality, that is $\|A-I\| = \sup_{\|x\| = 1} |\|Ax\|-\|x\||$.