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I have the following identity which I want to prove:

$$\frac{d}{dt} det(A+tB)|_{t=0} = Tr( Cof(A)^TB)$$

where $Cof(A)$ is the cofactor matrix of $A$, and $A$ is an $n\times n$ complex matrix. The reference is exercise 2 in section 1 of chapter 1 of Basic theory, volume I of PDE by Michael Taylor's book (2011 edition).

I tried the definition, i.e:

$$\frac{d}{dt}\det(A+tB)|_{t=0} = \frac{d}{dt}|_{t=0} \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^n (a_{i\sigma(i)}+tb_{i\sigma(i)})=$$ $$= \sum_{\sigma \in S_n } sgn(\sigma) \bigg[ b_{1\sigma(1)}\prod_{i=2}^n a_{i\sigma(i)}+\ldots + b_{n\sigma(n)} \prod_{i=1}^{n-1} a_{i\sigma(i)}\bigg]$$

How to proceed, I am not sure how does the cofactor of $A$ will look like in the notation with permutations, otherwise what is the missing step in the proof?

Thanks in advance.

1 Answers1

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Hint: You just need to apply this formula:https://en.wikipedia.org/wiki/Jacobi%27s_formula#Derivation

So you have $f(t)=A+tB$, $g(A)=det(A)$ so $f'(t)=B$, $(g\circ f)'(t)=Tr(Cof(A)^TB)$