Directional derivative of $x \mapsto A(x):= x_1 A_1 + x_2 A_2 + \dots x_n A_n $

Question

Let $\mathcal{S}_{m \times m}$ denote the space of real valued symmetric $m \times m$ matrices. Suppose $A_1, A_2 , \dots, A_n \in \mathcal{S}_{m \times m}$ are such symmetric $m \times m$-matrices.Now consider $A: \mathbb{R}^{n} \to \mathcal{S}_{m \times m}$ given by $x \mapsto A(x):= x_1 A_1 + x_2 A_2 + \dots x_n A_n \,.$ I am looking for the directional derivative $A^{\prime}(x;d)$ of $A$ in $x \in \mathbb{R}^{n}$ in the direction of $d \in \mathbb{R}^{n}$. Apparently $A$ is linear and therefore differentiable. Therefore I think that $A^{\prime}(x;d)$ can be expressed in thems of the derivative $A^\prime(x)$ but I have no clue what this could be...Any help is much appreciated.

Answer 1

For functions $f:\mathbb{R}^n \rightarrow \mathbb{R}$ the directional derivative in the direction $d$ is given by \begin{equation} f'(x;d) = \langle \nabla f(x), d \rangle = \sum_{i=1}^n \left ( \frac{\partial}{\partial x_i} f(x)\right )d_i \end{equation} The latter formula still holds even when you replace $f$, a scalar function, by $A:\mathbb{R}^n \rightarrow S_{m\times m}$, a matrix valued function.

In your case, the partials are \begin{equation} \frac{\partial}{\partial x_i}A(x) = A_i. \end{equation} Plugging this into the formula above gives \begin{equation} A'(x;d) = \sum_{i=1}^n \left ( \frac{\partial}{\partial x_i} A(x)\right )d_i = \sum_{i=1}^n (A_i) d_i. \end{equation} This should spiritually feel right since it's the same form as the directional derivative for scalar functions, except for the fact that the $A_i$ are matrices instead of scalars.