How to show $\frac{d}{dt}(\det(I+At)=\text{Tr}(A)$?

Question

How to show $\frac{d}{dt}\det(I+At)\mid_{t=0}=\DeclareMathOperator{\Tr}{Tr}\Tr(A)$? I can't find a easy way to prove that.

Did you mean that we should take the value of derivative at $0$? Otherwise this is not true. — zhoraster, Sep 11 '15 at 04:17
Unless I am mistaken, this is a special case of http://math.stackexchange.com/questions/202535/derivative-of-a-determinant-of-matrix. — Martin R, Sep 11 '15 at 04:22

score 2 · Accepted Answer · answered Sep 11 '15 at 04:29

You just need to look at the coefficient before $t$ in the polynomial obtained by expanding the determinant. Now try to identify where the terms $at$ come from. You should note that all the off-diagonal elements contain $t$, so there should not be too many such terms.

You can start with $2\times 2$ or $3\times 3$ to get the idea if you still stuck.

score -1 · Answer 2 · answered Sep 11 '15 at 04:27

-1

Hint: Note that the characteristic polynomial of $A$ is

$$\det(A-tI)=t^n-\mbox{Tr}(A)t^{n-1}+O(t^{n-2}) \; .$$

answered Sep 11 '15 at 04:27

Raskolnikov

16,108

Thank you, but what do you mean by O(t^{n-2})? – 6666 Sep 11 '15 at 04:28
1

That's really a terrible abuse of notation, since this is correct for unordered fields. Don't use $O(*)$ that way. – Thomas Andrews Sep 11 '15 at 04:29
@ThomasAndrews: Care to elaborate? – Raskolnikov Sep 11 '15 at 05:42

How to show $\frac{d}{dt}(\det(I+At)=\text{Tr}(A)$?

2 Answers2