Lipschitz constant of restricted determinant

Question

I'm trying to use optimization algorithms on the determinant restricted to a compact set $K$ and these algorithms require an explicit Lipschitz constant for the determinant.

For $n > 1$, the determinant is not a Lipschitz function. But, if we restrict the determinant to a compact set, the determinant is Lipschitz on that compact set $K$ because the determinant is $C^{\infty}(K)$.

I've tried to use the formula for the derivative of the determinant, but this involves the adjugate, which just leads us back the original problem. I'm not sure what else to try. If it helps, I can find an M such that $\{||X|| \leq M\colon X\in K\}$.

Answer 1

Using the derivative of $\det A$ (as here) is the way to go. As $h\to 0$, $$ \det(A+hB) = \det(A) + h\det(A)\operatorname{tr}(A^{-1}B) + O(h^2) = \det(A) +h\operatorname{tr}((\operatorname{cof} A)^T B) + O(h^2) $$ where $\operatorname{cof} A$ is the matrix of cofactors. Recall that $\operatorname{tr}(P^TQ)$ is the Euclidean inner product of matrices $P, Q$, i.e., $\sum_{ij} P_{ij}Q_{ij}$. This is naturally estimated by the Frobenius norm: $\operatorname{tr}(P^TQ) \le \|P\|_F \|Q\|_F$, where equality holds iff $Q$ is a scalar multiple of $P$. Conclusion so far: with respect to the Frobenius matrix norm, the linear functionsl $B\mapsto \operatorname{tr}((\operatorname{cof} A)^T B)$ has the norm $\|\operatorname{cof} A\|_F$.

Each entry of $\operatorname{cof}A$ is $\pm$ determinant of some $(n-1)\times (n-1)$ submatrix $C$, and we have $|\det C|\le \|C\|_F^{n-1}$ (geometric proof: the image of a unit cube under $C$ is a parallelepiped with sides at most $\|C\|_F$, hence with the volume of at most $\|C\|_F^{n-1}$). Hence, each entry of $\operatorname{cof} A$ is bounded by $\|A\|_F^{n-1}$, which implies $$ \|\operatorname{cof} A\|_F \le \sqrt{n^2\|A\|_F^{2(n-1)}} = n\|A\|_F^{n-1} $$ Conclusion: as measured in the Frobenius norm, $\|\nabla \det A\|\le n\|A\|_F^{n-1}$.