# Prove that if $f$ is linear, then for any $\textbf{a},\textbf{v} \in \mathbb{R}^2$, $f(\textbf{a}+\textbf{v})=f(\textbf{a})+[Df(\textbf{a})]\vec{v}$

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a function. Prove that if $f$ is linear, then for any $\textbf{a},\textbf{v} \in \mathbb{R}^2$, $f(\textbf{a}+\textbf{v})=f(\textbf{a})+[Df(\textbf{a})]\vec{v}$. Here $[Df(\textbf{a})]$ is the Jacobian matrix of $f$ evaluated at the point $\textbf{a}$. Thus, $[Df(\textbf{a})]\vec{v}$ is the directional derivative of $f$ at $\textbf{a}$ in the direction of $\vec{v}$

My attempt

Since $f$ is linear, $$f(\textbf{a}+\textbf{v})=f(\textbf{a})+f(\textbf{v}).$$

We need to show that $f(\textbf{v})=[Df(\textbf{a})]\vec{v}$. If $f=\begin{pmatrix} g_1 \\ g_2 \end{pmatrix}$, then $$[Df(\textbf{a})]=\begin{pmatrix} D_1 g_1 & D_2g_1 \\ D_1 g_2 & D_2 g_2\end{pmatrix}$$ Where $D_i f$ is the derivative of $f$ with respect to the $i^{th}$ variable with the rest of the variables held constant. Since $f$ is linear, we should have $D_1 g_2$ and $D_2g_1$ both zero (not sure how to prove this but it seems to make sense if we consider $g_1$ and $g_2$ to be the $x$ and $y$ co-ordinates, respectively. Thus: \begin{align} [Df(\textbf{a})]\vec{v}&= \begin{pmatrix} D_1 g_1 & 0 \\ 0 & D_2 g_2\end{pmatrix}\vec{v} \\ \\ &=\begin{pmatrix} D_1g_1(x_1) v_1 \\ D_2 g_2(x_2) v_2 \end{pmatrix}_{(x_1,x_2)=(a_1,a_2)} \end{align}

Where $\vec{v}=\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$. I'm not sure how to proceed from here.

Actually $D_1 g_2$ and $D_2 g_1$ are not zero.
The key observation is that $Df$ is constant (i.e. independent of $x$) and hence, higher derivatives are zero.
Oh, you may also consider the differential quotient $(f(a) - f(a+tv))/t$, use linearity and send $t$ to zero…
• I can do it through the differential quotient but how do i show $Df$ is constant? It seems trivial but i can't seem to see it Sep 11 '14 at 21:09
Using definition of derivative we see that for linear maps $Df_a=f$, (in the $\lim$ there $A_x(h)$ becomes $f(v)$, and the $\lim$ itself not needed). Let us now choose bases in the two spaces and find the (Jacobian) matrix of the linear map $f$. $f(v)=f(v^1e_1+v^2e_2)=f(e_1)v^1+f(e_2)v^2 = \bigl( \begin{smallmatrix} f(e_1) & f(e_2) \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} v^1 \\ v^2 \end{smallmatrix} \bigr)=Av,$ where the columns of A are the vectors $f(e_i)$ in the chosen image basis and independent of any particular $a$.