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Consider $\textbf{A} = exp(t\textbf{X})$ where $t \in \mathbb{R}$, $\textbf{X}, \textbf{A} \in \mathbb{R}^{n\times n}$ and $exp: \mathbb{R}^{n\times n} \rightarrow \mathbb{R}^{n\times n}$ is the matrix exponential. If we restrict $\textbf{X}$ to real-valued symmetric matrices, such that $\textbf{X}$ is diagonalizable and and $\textbf{A}$ is symmetric positive definite, is there a closed form of $\frac{d \textbf{A}}{d \textbf{X}}$?

Note that I am aware of the question here (Derivative of matrix exponential w.r.t. to each element of the matrix) but I believe my more restricted case of symmetric $\textbf{X}$ may yield a simpler and more concise solution.

Atkrye
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1 Answers1

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$ \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\vc#1{\op{vec}\LR{#1}} \def\Diag#1{\op{Diag}\LR{#1}} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} $A scalar function can be applied to a square matrix $$\eqalign{ f(x) = e^{tx} \qiq A &= f(X) \\ }$$ and a real symmetric matrix can be diagonalized with orthogonal factors $$\eqalign{ X &= QBQ^T,\quad B=\Diag b,\quad Q^TQ=I \\ }$$ The Daleckii-Krein theorem followed by vectorization yields $$\eqalign{ Q^T dA\:Q &= R\odot\LR{Q^T dX\:Q} \\ \LR{Q\otimes Q}^T\,da &= \vc{R}\odot\LR{Q\otimes Q}^Tdx \\ }$$ where $(\odot,\otimes)$ denote the Hadamard and Kronecker products, respectively.

By defining the matrices $$\eqalign{ D &= \Diag{\vc{R}},\qquad P = \LR{Q\otimes Q}\qiq P^T=P^{-1} \\ }$$ the gradient calculation simplifies to $$\eqalign{ P^T da &= DP^Tdx \qiq \grad{a}{x} = PDP^T \\ }$$ All that remains is to calculate the symmetric $R$ matrix which lies at the heart of the theorem $$\eqalign{ R_{ij} = \begin{cases} {\Large\frac{f(b_i)-f(b_j)}{b_i-b_j}} \qquad{\rm if}\;b_i\ne b_j \\ \\ \quad f'(b_j) \qquad\qquad {\rm otherwise} \\ \end{cases} }$$ NB:$\:$ For the current problem $\;f'(x) = te^{tx}$

greg
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