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Let $f:A\mapsto \rm{det}(A)$, Prove that $\left(Df\right)_{{\rm id}}\left(H\right)={\rm tr}\left(H\right)$ for all $H\in\mathcal{L}\left(\mathbb{R}^{n}\to\mathbb{R}^{n}\right)$.

The question appears also here: Directional derivative of the determinant but with no answers apart from that of the poster itself, and his solution uses some identities regarding the characteristic polynomial I do not understand. Additionally I think his approach assumes $H$ is invertible, which we are not given.

Naturally I have also tried calculating the directional derivative, giving

$$ \lim_{t\to0}\frac{\det(tH+\rm{id})-\det{\rm{id}}}{t}=\lim_{t\to0}\frac{\det(tH+\rm{id})-1}{t} $$

where I want to prove it inductively by extracting the first row to have

$$\rm{det}(tH+\rm{id}) = \rm{det}\pmatrix{1\ 0\ \dots \ 0 \\(tH+\rm{id})_2\\\ \vdots \ \\(tH+\rm{id})_n} + t\rm{det}\pmatrix{H_{1,1}\ H_{1,2}\ \dots \ H_{1,n} \\(tH+\rm{id})_2\\\ \vdots \ \\(tH+\rm{id})_n} $$

(where $(A)_i$ is just the $i$th row of the matrix) where I want to say that the left determinant is the sum of $tH_{i,i}$ for $i\geq 2$ by induction which leaves me with showing the right side is $tH_{1,1}$, but I'm not sure how to proceed with that.

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Nescio
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1 Answers1

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If you use the definition of determinant (sum over the permutation) you get $$ \det(I+t H)=1+t\, {\mbox tr} H+o(t) $$ (check that the identity permutation is the one giving all terms of order $0$ and $1$ in $t$: for any other permutation you will get at least $t^2$). This readily implies what you want.

John B
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