Derivative of the determinant of the right stretch tensor

Question

I have to evaluate the derivative $$ \frac{\partial\det\mathcal{U}}{\partial F} $$ where $\mathcal{U}=\sqrt{F^TF}$ and $F$ is a $m\times n$ real matrix. Any suggestion would be appreciated.

Thank you all, guys!! You helped me a lot.

Answer 1

You can also approach the problem using differentials instead of the chain rule. It is easy to work with differentials, because algebraically they act like ordinary matrices.

Define a new matrix variable $W$, and its differential $$\eqalign{ W &= U^2 = F^TF = W^T \cr dW &= 2\,\operatorname{sym}(F^T\,dF)\cr }$$where $\operatorname{sym}(A)=\frac{1}{2}\,(A+A^T)$ is the symmetrization operation.

Now you want to find the gradient of $$\eqalign{ g &= \det(U) = \sqrt{\det W} \cr }$$ but it's more convenient work with its logarithm instead $$\eqalign{ h &= \log(g) = \frac{1}{2}\,\log\det W\cr dh &= d\log(g) = \frac{1}{2}\,d\operatorname{tr}\log W \cr \frac{dg}{g} &= \frac{1}{2}\,W^{-1}:dW \cr dg &= \frac{g}{2}\,W^{-1}:dW \cr &= g\,W^{-1}:\operatorname{sym}(F^T\,dF) \cr &= g\,\operatorname{sym}(W^{-1}):F^T\,dF \cr &= g\,W^{-1}:F^T\,dF \cr &= g\,FW^{-1}:dF \cr \cr }$$ where colon denotes the double-dot (aka Frobenius) product, which can be defined as $$A:B=\operatorname{tr}(A^TB)$$

So the gradient of interest is $$\eqalign{ \frac{\partial g}{\partial F} &= gFW^{-1} \cr &= F(F^TF)^{-1}\det\sqrt{F^TF} \cr }$$

Answer 2

Hint.

Name $f(F) = \det \sqrt{F^T F}$ You can write $f= f_3 \circ f_2 \circ f_1$ as the function composition of

$$\begin{array}{l|rcl} f_1 : & \mathcal M_{m,n}(\mathbb R) & \longrightarrow & \mathcal S^+(\mathbb R^n) \subseteq \mathcal M_{n}(\mathbb R) \\ & F & \longmapsto & F^T F \end{array}$$

$$\begin{array}{l|rcl} f_2 : & \mathcal S^+(\mathbb R^n) & \longrightarrow & \mathcal S^+(\mathbb R^n) \\ & M & \longmapsto & \sqrt{M} \end{array}$$

$$\begin{array}{l|rcl} f_3 : & \mathcal M_{n} & \longrightarrow & \mathbb R \\ & M & \longmapsto & \det M \end{array}$$

And finally

$$\begin{array}{l|rcl} f : & \mathcal M_{m,n}(\mathbb R) & \longrightarrow & \mathbb R \\ & F & \longmapsto & \det \sqrt{F^T F} \end{array}$$

You want to compute the Fréchet derivative $f^\prime$ of $f$. Applying the chain rule twice, you have: $$f^\prime(F)=(f_3^\prime((f_2 \circ f_1)(F))) \circ f_2^\prime(f_1(F)) \circ f_1^\prime(F)$$

Now you have to find the Fréchet derivative of $f_1$, $f_2$ and $f_3$.

You have (Jacobi Formula): $$f_3^\prime(F).H = \text{tr}(\text{adj}(F)H)$$ and (based on the derivative of a bilinear function between vector spaces)

$$f_1^\prime(F).H=F^T H + H^T F$$

The most difficult is to compute $f_2^\prime$. That can be done using Implicit function theorem, knowing that $$(f_2(F))^2=F^T F$$ See Derivative (or differential) of symmetric square root of a matrix for more details on that last point.