How do I show that the derivative of the path $\det(I + tA)$ at $t = 0$ is the trace of $A$? Here, I'm taking $A$ to be a linear operator on $\mathbb R^n$ for $n>1$. Can you please tell me how to solve such a problem?
 A: We have
$$\det(I+tA)=t^n\det\left(A+\frac1t I\right)=t^n\chi_A\left(\underbrace{-\frac 1t}_{=X}\right)=t^n\left((-1)^nX^n+(-1)^{n-1}\operatorname{tr}(A) X^{n-1}+\cdots+\det A\right)\\=1+ \operatorname{tr}(A)t+\cdots+\det(A)t^n$$
and the result follows.
A: Letting $B=I+t A$, and using the Leibniz formula of the determinant, in terms of the permutations of the set $\{1,2, \ldots, n\}$:
$$p(t)=\det(I+tA)=\det(B)=\sum_\sigma {\rm sign}(\sigma) \prod_{i=1}^n b_{i,\sigma_i} \tag{1}$$
One of the terms of the sum consists of the permutation that picks the diagonal elements; this has the form:
$$(t\, a_{1,1}+1)(t\, a_{2,2}+1)\cdots (t\, a_{n,n}+1)=1+ \beta_1 t + \beta_2 t^2+\beta_3 t^3 + \cdots + \beta_n t^n$$
where $\beta_1 = a_{1,1}+a_{2,2}+\cdots + a_{n,n}={\rm tr}(A)$
The other terms in $(1)$ include at least two off-diagonal elements in the product, hence they include a factor $t^2$.
Hence, $p(t)=1+ {\rm tr}(A)\,t + t^2 g(t)$ where $g(t)$ is some polynomial; and the result follows.
A: we will use the following facts:
(a) determinant of a matrix is the product of the eigenvalues,
(b) trace of a matrix is the sum of the eigenvalues,
(c) eigenvalues of $I + tA$ are $1 + t \lambda$ where $\lambda$ is an eigenvalue of $A.$ 
$\det(I + tA) = (1 + t\lambda_1)(1 + \lambda_2)\cdots = 1 + t(\lambda_1 + \lambda_2+\cdots) + \cdots = 1 + \operatorname{trace}(A) t + \cdots$
therefore, the derivative of $\det(I + tA)$ at $t = 0$ is the $\operatorname{trace}(A).$ 
A: For small $t$, we have $\det (\exp tA) = \exp (\operatorname{tr} tA) = \exp (t \operatorname{tr} A) = 1 + (\operatorname{tr} A)t + O(t^2)$. (If the first equality isn't clear, use the fact that $\exp$ and $\det$ behave nicely under conjugation, and reduce the statement to Jordan normal form.)
