Derivative of matrix-valued function with respect to scalar

Question

Given vector $\mu \in \Bbb R^n$ and $n \times n$ matrices $A$ and $\Sigma$, let matrix-valued function $F : \Bbb R \to \Bbb R^{n \times n}$ be defined by

$$F(t) := |(I-2tA\Sigma)|^{1/2} \exp \left(\frac{1}{2}\mu'[I-(I-2tA\Sigma)^{-1})\Sigma^{-1}\mu \right)$$

How can I get the second derivative of $F$ with respect to $t$? Could someone please give me a hint?

Answer 1

Try matrixcalculus.org. Entering

det(eye-2*t*A*S)^(1/2)*exp((1/2) * v'*(eye-inv(eye-2*t*A*S)*inv(S))*v)

yields

function: $$ f = \mathrm{det}(\mathbb{I}-2\cdot t\cdot A\cdot S)^{(1/2)}\cdot \exp(1/2\cdot v^\top \cdot (\mathbb{I}-\mathrm{inv}(\mathbb{I}-2\cdot t\cdot A\cdot S)\cdot \mathrm{inv}(S))\cdot v) $$

gradient: $$\begin{aligned} \frac{\partial f}{\partial t} = &-(\mathrm{det}(\mathbb{I}-2\cdot t\cdot A\cdot S)^{(1/2-1)}\cdot \exp((v^\top \cdot (\mathbb{I}-\mathrm{inv}(\mathbb{I}-2\cdot t\cdot A\cdot S)\cdot \mathrm{inv}(S))\cdot v)/2)\cdot \mathrm{tr}(A\cdot S\cdot \mathrm{adj}(\mathbb{I}-2\cdot t\cdot A\cdot S)) \\&+\mathrm{det}(\mathbb{I}-2\cdot t\cdot A\cdot S)^{(1/2)}\cdot \exp((v^\top \cdot (\mathbb{I}-\mathrm{inv}(\mathbb{I}-2\cdot t\cdot A\cdot S)\cdot \mathrm{inv}(S))\cdot v)/2)\cdot v^\top \cdot \mathrm{inv}(\mathbb{I}-2\cdot t\cdot A\cdot S)\cdot A\cdot S\cdot \mathrm{inv}(\mathbb{I}-2\cdot t\cdot A\cdot S)\cdot \mathrm{inv}(S)\cdot v) \end{aligned}$$

You can try plugging this result back into the engine.

Answer 2

For typing convenience, define the matrices $$\eqalign{ W &= \mu\mu^T,\qquad&S &= \Sigma^T \\ B &= I - 2tAS^T,\qquad &M &= B^{-1}AS^T\\ \dot B &= -2AS^T = -2BM,\qquad&\dot M &= -B^{-1}\dot BB^{-1}AS^T = 2M^2 \\ \dot B^{-1} &= -B^{-1}\dot BB^{-1} = 2MB^{-1} \\ }$$ and the scalars $$\eqalign{ X &= \tfrac 12WS^{-1}:(I-B^{-1}) \\ \dot X &= \tfrac 12WS^{-1}:(-\dot B^{-1}) = -WS^{-1}:MB^{-1} \\ \\ E &= \exp(X) \\ \dot E &= E \dot X \\ \\ F &= E\;(\det B)^{1/2} \\ L &= \log F = \log(E) + \tfrac 12\log(\det B) = X + \tfrac 12{\rm Tr}(\log(B)) \\ \dot L &= \dot X + \tfrac 12 B^{-T}:\dot B \;=\; -(WS^{-1} + B^T):MB^{-1} \\ }$$ So the first derivative of $F$ is $$\eqalign{ \def\FF{\left(\frac{\dot F}{F}\right)} \def\B{B^{-1}} \def\DB{\dot B^{-1}} \def\S{S^{-1}} \def\a{\alpha} \dot L &= \FF \quad&\implies &\dot F = -F\;(W\S + B^T):M\B \\ \dot F &= F\a &\implies\quad &\a = \FF = -(W\S + B^T):M\B \\ }$$ Now the second derivative of $F$ can be calculated. $$\eqalign{ \ddot F &= \dot F\a + F\dot\a \\ \\ \frac{\ddot F}{F} &= \FF\a + \dot\a \\ &= \a^2 - (W\S+B^T):(\dot M\B+M\DB) \;-\; (\dot B)^T:(M\B) \\ &= \a^2 - 2(W\S + B^T):(M^2\B+M^2\B) \;+\; 2(BM)^T:(M\B) \\ &= \a^2 - 4(W\S + B^T):(M^2\B) \;+\; 2B^T:(M^2\B) \\ &= \a^2 - \Big(4W\S+2B^T\Big):\Big(M^2\B\Big) \\\\ }$$ In several of the steps above, a colon is used to denote the trace/Frobenius product, i.e. $$\eqalign{ A:B = {\rm Tr}(A^TB) = {\rm Tr}(B^TA) = B:A }$$