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Let $A\in\mathbb{R}^{n\times n}$ be an invertible matrix, $T\in\mathbb{R}^{n\times n}$ arbitrary.

Show $$(\det)'(A)(T)=\det(A)\operatorname{tr}(A^{-1}T)$$

Fist I need to determine $(\det)'(I)(T)=\lim_{\epsilon\to 0}\frac{\det(I+\epsilon T)-\det(I)}{\epsilon}$ and then use chain rule.

I have found many identical questions and a similar proof on Wikipedia https://en.wikipedia.org/wiki/Jacobi%27s_formula#Derivation (via chain rule) but I don't understand them. Why is $\lim_{\epsilon\to 0}\frac{\det(I+\epsilon T)-\det(I)}{\epsilon} = \operatorname{tr}(T)$? Thank you!

user1049882
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    See the second half of my answer here. The approach there uses the existence of an upper-triangular matrix which is similar to $T$, but over the field $\Bbb{C}$ (e.g the Jordan canonical form). Alternatively, you can just write out the determinant using one of the explicit formulae, e.g the one involving permutations, or using induction and cofactor expansion (it's too messy for my taste, which is why I didn't go elaborate on that). – peek-a-boo Jun 16 '22 at 16:27
  • @peek-a-boo I have a question to your second half: the field here is $\mathbb{R}$ and not $\mathbb{C}$, how do we know that $T$ is triangularizable? – user1049882 Jun 16 '22 at 16:52
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    every real matrix can be considered a complex matrix. So imagine your $T$ as living in $M_{n\times n}(\Bbb{C})$, and perform the triangularization there. Even more explicitly, $\det_{\Bbb{R}}(I+\epsilon T)=\det_{\Bbb{C}}(I+\epsilon T)=\det_{\Bbb{C}}(I+\epsilon PUP^{-1})=\det_{\Bbb{C}}(P(I+\epsilon U)P^{-1})=\det_{\Bbb{C}}(I+\epsilon U)=\text{tr}{\Bbb{C}}(U)=\text{tr}{\Bbb{C}}(T)=\text{tr}_{\Bbb{R}}(T)$, where the subscripts indicate which field I'm imagining the matrices to live in. – peek-a-boo Jun 16 '22 at 16:53
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    ^oops I forgot to write $\frac{d}{d\epsilon}\bigg|{\epsilon=0}$ at each step above (the equalities above are mostly right, the only erroneous one is $\det{\Bbb{C}}(I+\epsilon U)=\text{tr}_{\Bbb{C}}(U)$; it is the derivative of the left side which equals the right). – peek-a-boo Jun 16 '22 at 17:15

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The key is to consider the polynomial $\det(I + \epsilon T)$. One way to get its linear term is to consider its relation to the characteristic polynomial. In particular, we have $$ \det(I + \epsilon T) = \det(\epsilon \cdot [\epsilon^{-1} I -(-T)]) = \epsilon^n \det(\epsilon^{-1} I - (-T)). $$ If the characteristic polynomial of $-T$ is $p(t) = \det(tI - (-T))$, then $\det(I + \epsilon T) = \epsilon^n p(\epsilon^{-1})$. Notably, this is simply the polynomial whose coefficients are the coefficients of $p$ in reverse order. Because the coefficient of $t^{n-1}$ in $p(t)$ is $\operatorname{tr}(T)$, it follows that the coefficient of $\epsilon^1$ in $\epsilon^{n}p(\epsilon^{-1})$ is also $\operatorname{tr}(T)$.

Alternatively, you can derive this fact directly using the Leibniz formula for the determinant $\det(I + \epsilon T)$.

Ben Grossmann
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