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I want to show that if for $f: \mathbb{R}^2 \mapsto \mathbb{R}^2$ if we have

$f\begin{pmatrix} a_1 + v_1 \\ a_2 + v_2 \\ \end{pmatrix}$ = $f\begin{pmatrix} a_1 \\ a_2 \\ \end{pmatrix} + [Df(a)] \large\vec{v}$

for all $a,v \in \mathbb{R}^2$

then $f$ is linear.

Obviously the derivative is linear, not sure how that can help me here though, since I can't see any immediate use for the equality $[Df(a)] \large\vec{v}$ $=$ $[Df(a\vec{ v})]$

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  • $\begingroup$ One approximates a differentiable function by a linear function; what happens if that function equals this approximation? Answer: It is linear. :) Hint: Try to verify the linearity axioms. $\endgroup$
    – awllower
    Mar 10 '14 at 13:36
  • $\begingroup$ Well, that's akin to showing $D[f(a)]\vec{v} = f(\begin{matrix} v_1 \\ v_2 \end{matrix}$ And to be honest, I'm not sure how to do that without f being linear.. $\endgroup$ Mar 10 '14 at 19:22
  • $\begingroup$ in the first line there may be a typo: should be $\to$ rather than $\mapsto$ $\endgroup$
    – rych
    Mar 11 '14 at 7:41
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You have $\forall x\in\mathbb{R}^2,f(x)=f(0)+Df(0)x$ which looks linear to me.

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  • $\begingroup$ is showing only 0 work slike that sufficient to show its linear? $\endgroup$ Mar 11 '14 at 19:35

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