Giving $\nabla^{2} f(\mathbf{x}) \succeq mI$,how to calculate $\|(\nabla^{2} f(\mathbf{x}))^{-1}\| $

Question

In the 《Introduction to Nonlinear Optimization Theory, Algorithms, and Applications with MATLAB》page 85, "Combining the latter equality with the fact that $\nabla^{2} f(\mathbf{x}) \succeq mI$ implies that $\|(\nabla^{2} f(\mathbf{x}))^{-1}\| \leq \frac{1}{m}$",how to deduce.$I$ is identity matrix and $\succeq$ is positive semidefinite.

Answer 1

The Hessian $\nabla^2 f(x)$ is symmetric and positive definite, hence there exists an orthogal matrix $Q$ and a diagonal matrix $D$ such that $$ \nabla^2 f(x)=QDQ^T $$ We can then proceed the check that $\nabla^2 f(x)-mI=QDQ^T-mQQ^T=Q(D-mI)Q^T \geq 0$ implies $$ D-mI \geq 0 $$ Hence no entry of the diagonal matrix $D$ is smaller than $m$. However, the inverse of $D=diag(d_1,...,d_n)$ is just given by $D^{-1}=diag(d_1^{-1},...,d_n^{-1})$ (note that no entry is bigger than $\frac{1}{m}$) and the inverse of $\nabla^2 f(x)$ is given by $$ (\nabla^2 f(x))^{-1}=QD^{-1}Q^T $$ We can now proceed to compute its norm: $$ ||(\nabla^2 f(x))^{-1}||_2=\sup_{||y||_2=1}||(\nabla^2 f(x))^{-1}y||_2= \\ \sup_{||y||_2=1}||QD^{-1}Q^Ty||_2=\sup_{||y||_2=1}||D^{-1}y||_2 \leq \frac{1}{m} $$ There are different proofs using the properties of the matrix norms, but I went with a more elementary approach.