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Let $f: \mathbb{R}^p \rightarrow \mathbb{R}^q$ be linear.

We have proven that for $T$ linear it is: $$ D T(a) = T. $$

So it should imply that: $$ D(Df(a))(a) = f. $$ Right? This seems to be trivial, but I had a long discussion with a fellow student who keeps saying that it is $0$ and also, that considering such examples is absolutely irrelevant in mathematical praxis.

So who is wrong...?

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  • $\begingroup$ @CameronWilliams I think the OP's notation is OK. In this usage $Df(a)$ is not a vector in $\mathbb R^q,$ it's the linear map that approximates the map $f-f(a)$ near $a.$ $\endgroup$
    – zhw.
    Commented Jun 23, 2021 at 16:03
  • $\begingroup$ You are right, it is trivial. Your fellow student would do well to understand why. $\endgroup$
    – zhw.
    Commented Jun 23, 2021 at 16:05
  • $\begingroup$ I dont know the notation $DT$, we only (yet) defined the total derivative of a function in a point. So I only know $DT(a)$ $\endgroup$
    – mathematics-and-caffeine
    Commented Jun 23, 2021 at 16:07

1 Answer 1

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You have to be careful what you're taking the derivative of. $\mathrm Df(a):x\mapsto \mathrm Df(a)(x)$ is a linear map for fixed $a$. However, the map $\mathrm Df:a\mapsto\mathrm Df(a)$ is a (usually) nonlinear map assigning to each $a$ a linear map.

Differentiating the first one will return the same map, since it's linear, and the derivative of a linear map is exactly that linear map. However, differentiating the second one will give you something different, and it's what we usually care about.

In the first case, if $f$ is linear, then you will indeed get $\mathrm D(\mathrm Df(a))(b)=f$ for all $a$ and $b$. In the second case, since $\mathrm Df(a)$ is a constant, you will get $\mathrm D(\mathrm Df)(a)=0$. Notice the meaningful difference in notation. The way you wrote it, we should actually mean the first case Your friend is right in that this is not a case you'll often see discussed, though.

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    $\begingroup$ Note the hypothesis that $f:\mathbb R^p\to\mathbb R^q$ is linear, so $Df:\mathbb R^p\times\mathbb R^p\to\mathbb R^q$ maps $Df(a)(b)=f(b)$. Thus, for each $a\in\mathbb R^p$, $Df(a)=f$. (unless you want to make a distinction between the space and its tangent space, which is not usually done at this stage). $\endgroup$
    – Jackozee Hakkiuz
    Commented Jun 23, 2021 at 16:01
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    $\begingroup$ Nice answer (+1). Maybe you could mention that in the second case the second derivative is indeed the zero function? So in a sense both of the people mentioned in the post are right :) $\endgroup$
    – Severin Schraven
    Commented Jun 23, 2021 at 16:04
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    $\begingroup$ Yes, it's a linear function. But it's always the same linear function, independent of $a$, assuming that $f$ itself is linear. We have a function which maps vectors to functions (which are also vectors in an appropriate vector space!), and as such it is a constant function, because it always maps to the same function. $\endgroup$
    – Vercassivelaunos
    Commented Jun 24, 2021 at 6:19
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    $\begingroup$ @LukasNiessen Yes, exactly. $\endgroup$
    – Vercassivelaunos
    Commented Jun 24, 2021 at 15:21
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    $\begingroup$ @LukasNiessen Yes, exactly. $\endgroup$
    – Vercassivelaunos
    Commented Jul 9, 2021 at 16:36

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