TITLE: How do I show that the derivative of the path $\det(I + tA)$ at $t = 0$ is the trace of $A$?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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 $\operatorname{trace}(A).$