# Matrix derivative w.r.t. a general inverse form: $(A^TA)^{-1/2}D(A^TA)^{-1/2}$

I want to find derivative of matrix $$(A^TA)^{-1/2}D(A^TA)^{-1/2}$$ w.r.t. $$A_{ij}$$ where D is a diagonal matrix. Alternatively, it is okay too to have

$$\frac{\partial}{\partial A_{ij}} a^T(A^TA)^{-1/2}D(A^TA)^{-1/2}b$$

Is there any reference for such problem? I have the matrix cookbook which gives results when $$D=I$$. But how is this general form evaluating to?

To give more information, empirical distribution of diagonal of diagonal matrix D converges to some known distribution.

• try matrixcalculus.org ... Jun 11, 2021 at 23:55
• I'm afraid it does not support square root of matrix. Jun 12, 2021 at 0:05

Since calculating the derivative of $$B=(A^T A)^{-1}$$ with respect to $$A$$ is familiar, $$dB=-(A^TA)^{-1}(dA^TA+A^TdA)(A^TA)^{-1},$$ let me compute the derivatives with respect to $$B$$, to focus on the square root difficulties. So we seek the derivative of $$X=\sqrt{B}D\sqrt{B}$$ with respect to the symmetric positive definite matrix $$B$$. Start from $$dX=(d\sqrt{B})D\sqrt B+\sqrt{B}D(d\sqrt{B}).$$ Vectorize, $$\text{vec}\,dX=\bigl[\sqrt{B} D\otimes I+I\otimes \sqrt{B}D\bigr]\,\text{vec}\,(d\sqrt{B}),$$ and use the identity $$\text{vec}\,(d\sqrt{B})=(\sqrt{B}\oplus\sqrt{B})^{-1}\,\text{vec}\,dB,$$ to arrive at $$\text{vec}\,d(\sqrt{B}D\sqrt{B})=\bigl[\sqrt{B} D\otimes I+I\otimes \sqrt{B}D\bigr]\,(\sqrt{B}\oplus\sqrt{B})^{-1}\,\text{vec}\,dB.$$ Here $$\otimes$$ and $$\oplus$$ are Kronecker products and sums.