How do I prove that $f'(x_0) =\text{tr}(A)$ where $x_0 = 0$ and $f(x) = \det(E+Ax)$ for some square matrix $A$?
P.S. I'm pretty sure there exists a solution that doesn't use anything more complicated than definitions of the determinant, trace and derivative. I'm just starting to learn linear algebra and I'm only familiar with basic things.
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I work in the field $C$, let $P$ be a matrix such that $PAP^{-1}$ is sup triangular, $c_1,...c_n$ the eigenvalues of $A$, $det(E+xA)=det(P(E+xA)P^{-1})=det(E+xPAP^{-1})=(1+xc_1)...(1+xc_n)$, this implies that $det(E+xA)=1+x(c_1+...+c_n)+x^2U(x)$ and henceforth ${d\over{dx}}_{x=0}=c_1+...+c_n=tr(A)$.
To find the derivative of $\det(I+tA)$ at $t=0$, we want to write $$ \det(I+tA) = \det(I) + ct + o(t) $$ for some $c$, which will be the derivative.
If we expand $\det(I+tA)$ by the Leibniz formula, we find that almost all of the terms contain at least two factors of $t$ and so belong in the $o(t)$ part. The only one this isn't the case for is the product of the diagonal elements: $$ \det(I+tA) = (1+ta_{11})(1+ta_{22})\cdots(1+ta_{nn}) + \text{terms involving }t^2 $$
Expanding this product with the distributive law we get $$ \det(I+tA) = 1 + (ta_{11}+ta_{22}+\cdots+ta_{nn}) + \text{terms involving }t^2 + \text{more such terms} $$
The terms of degree 1 are exactly $t$ times the trace of $A$, so this trace must be the derivative we're looking for.
If you're not familiar with little-$o$ notation, you can also work with the high-school definition of derivative; we get $$ \lim_{t\to 0} \frac{1+(ta_{11}+\cdots+ta_{nn})+(\text{terms involving }t^2) - 1}{t-0} = a_{11}+\cdots+a_{nn}$$