Let A be a square, symmetric, positive-definite matrix. Let S be its symmetric square root found by a singular value decomposition. Let vech() be the half-vectorization operator.
Is there a convenient expression for the derivative (or differential) of vech(S) with respect to vech(A)?
I know the expression for an inverse, which is sort of like a matrix version of the power rule. Would this approach work for a symmetric square root as well (i.e., (1/2)S^(-1/2))?
Best Answer
What follows is an extension of the previous comments, to derive an explicit expression in terms of Kronecker sum. Taking differential $\mathrm{d}(\cdot)$ to both sides of $\sqrt{A}\sqrt{A} = A$ results a special case of Sylvester equation $$(\mathrm{d}\sqrt{A}) \sqrt{A} \: + \: \sqrt{A} (\mathrm{d}\sqrt{A}) = \mathrm{d}A, $$ which can be solved for the differential matrix $\mathrm{d}\sqrt{A}$ as $$ \text{vec}(\mathrm{d}\sqrt{A}) = \left(\sqrt{A}^{\top} \oplus \sqrt{A}\right)^{-1} \: \text{vec}(\mathrm{d}A). $$ Since $A$ is positive definite, $\sqrt{A}$ is unique and positive definite, and hence the Kronecker sum is positive definite (thus non-singular). Further, since the differential and vec operator can be interchanged in the left hand side of the equation above, the Jacobian identification rule (p. 198 in Magnus and Neudecker, Matrix Differential Calculus with Applications in Statistics and Econometrics, 3rd ed., chapter 9, section 5) results $$ \mathrm{D}\sqrt{A} = \left(\sqrt{A}^{\top} \oplus \sqrt{A}\right)^{-1}, $$ where the transpose can be dispensed if $A$, in addition to being positive-definite, is also symmetric, as the OP asked. Notice that for generic matrix function $F: \mathbb{R}^{p\times q} \mapsto \mathbb{R}^{m\times n}$, the Jacobian is defined as $\mathrm{D}F(X) \triangleq \displaystyle\frac{\partial \: \text{vec}(F(X))}{\partial \: (\text{vec}(X))^{\top}}$, and is of size $mn \times pq$.