Characteristic polynomial, trace, and determinant.

Let $V$ be a finite dimensional vector space over $F$ and $T$ be a linear operator on $V$. If $\beta$ an ordered basis of $V$, then we want to establish the following in this post:
Let $f$ be the characteristic polynomial of $T$, then

$f(t) = t^n -(tr(T))t^{n-1} + \cdots + (-1)^n det(T)$

Let's recall the definition of the characteristic polynomial of $T$, 

$f(t) = det(T - tI)$

where $I$ is the appropriate identity transformation. If $F$ is itself not algebraically closed, we consider $\overline{F}$, the algebraic closure of $F$, then $f(t)$ splits with the form,

$f(t) = \prod_{i=1}^n (t-\lambda_i)$

Since we are working on $\overline{F}$, we could find the Jordan Canonical form of $T$, hence in a matrix language $[T]_\beta = QAQ^{-1}$ where $A$ is a upper-triangular matrix with diagonal being eigenvalues.

Then $det(T) = det(QAQ^{-1}) = det(A)$ which is the product of the eigenvalues. (up to a sign) Similiarly, we have that $tr(T) = tr(A) = \sum \lambda_i$, hence coincide with the coefficient of $t^{n-1}$ form the characteristic polynomial.