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I want to prove differential of det map at some matrix $A$ is given by $f(A)\operatorname{tr}(X)$

let $f:GL_n(\mathbb{R})\rightarrow\mathbb{R}$ is det map i.e $f(A)=\det A$,

claim: $f_{*}(AX)=(f(A))\operatorname{tr}(X)$

proof of claim:

Let a curve $c(t)=Ae^{tX}$ such that $c(0)=A,c'(0)=AX$. using this curve we calculate differential

$f_{A{*}}(AX)=\frac{d}{dt}|_{t=0}f(c(t))=c'(t)|_{t=0}f(c(t))|_{t=0}=AXf(A)=AX(\det A)$, where am I getting wrong? Thank you.

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up vote 2 down vote accepted

It seems your problem is that you forgot to take the derivative (gradient) of the determinant $f:\mathbb R^{n^2}\to\mathbb R$. I.e., you should have $$\frac{d}{dt}f(c(t))|_{t=0} = \left[\nabla f(c(t))c'(t)\right]|_{t=0}$$

But there is maybe a better way, and it is worked out here.

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I want to do it in my way, I have no idea of $\nabla$ – Un Chien Andalou Aug 19 '12 at 3:53
@Flute, right, but my point is that you didn't properly apply the chain rule to differentiate! – Andrew Aug 19 '12 at 3:57
:-o :-o :-o :-o :-o – Un Chien Andalou Aug 19 '12 at 4:00
:-):-):-):-):-) – Andrew Aug 19 '12 at 4:01

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