Gradient of $X \mapsto \frac1{\sqrt{\det (X)}}$

Question

Let $\Bbb S_n^{++}$ denote the set of $n \times n$ symmetric positive definite matrices over $\Bbb R$. Let scalar field $f : \Bbb S_n^{++} \to \Bbb R$ be defined by $$ f (X) := \frac1{\sqrt{\det(X)}} $$ Find the gradient $\nabla_X f$.

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What I got was

$$\nabla_X f (X) = -\frac12\det(X)^{-\frac32} \operatorname{tr} \left( X^{-1} \right)$$

Is it correct? Full explanation, please.

Answer 1

The gradient of the determinent is well known $$\eqalign{ \def\qiq{\quad\implies\quad} \def\LR#1{\left(#1\right)} \def\p{\partial}\def\grad#1#2{\frac{\p #1}{\p #2}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} h &= \det(X) \qiq\grad hX = hX^{-T} \qquad\qquad\quad \\ }$$ Using this result, the gradient of your function becomes $$\eqalign{ f &= h^{-1/2} \qiq \grad fX &= -\frac{1}2\:h^{-3/2}\gradLR hX \\ &&= -\frac 12\:h^{-1/2}X^{-T} \\ }$$ The result that you derived is incorrect because it is scalar-valued, whereas a scalar-by-matrix gradient must be matrix-valued.