The Total Derivative of a Linear Function from Rn to Rm
Recall from the Differentiability and the Total Derivative of Functions from Rn to Rm page that if $S \subseteq \mathbb{R}^n$, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ then $\mathbf{f}$ is said to be differentiable at $\mathbf{c}$ if there exists a linear function $\mathbf{T}_{\mathbf{c}} : \mathbb{R}^n \to \mathbb{R}^m$ called the total derivative of $\mathbf{f}$ at $\mathbf{c}$ such that for all $\mathbf{v} \in \mathbb{R}^n$ we have that:
(1)Where $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$.
Now suppose that $\mathbf{f}$ is a linear function. Then the total derivative of $\mathbf{f}$ at any defined point $\mathbf{c}$ is extremely easy to determine as we will see in the following proposition.
Proposition 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$. If $\mathbf{f}$ is a linear function defined at $\mathbf{c}$ then $\mathbf{f}$ is differentiable at $\mathbf{c}$ and the total derivative of $\mathbf{f}$ at $\mathbf{c}$ is $\mathbf{f}$ itself. In other words, for all $\mathbf{v} \in \mathbb{R}^n$, $\mathbf{T}_{\mathbf{c}} (\mathbf{v}) = \mathbf{f}(\mathbf{v})$. |
- Proof: Let $\mathbf{f}$ be a linear function. Then for any $\mathbf{v} \in \mathbb{R}^n$ we have that:
- Set $\mathbf{T}_{\mathbf{c}}(\mathbf{v}) = \mathbf{f}(\mathbf{v})$ and $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) = \mathbf{0}$. Then we have that:
- Where $\mathbf{T}_{\mathbf{c}} = \mathbf{f}$ is a linear function and $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$. So $\mathbf{f}$ is differentiable at $\mathbf{c}$ and for all $\mathbf{v} \in \mathbb{R}^n$: