Determining the coefficients of the characteristic polynomial I am trying to understand the expansion of characteristic polynomials of $A \in  M_{n\times n}$ in the form 
$$ p(t) = (-1)^n t^n+\dots+a_1 t + \det(A),$$ but despite quite some time googling, I can't figure out what the $a_i$ coeffecients in $a_i t^i$ (terms in the middle) come from. Could anyone explain how to expand $p(t)=\det(A-tI)$ into this form?
Thanks in advance.
 A: Note: Often the characteristic polynomial of $A$ is defined as $p(t) = \det(tI - A)$ rather than $p(t) = \det(A -tI)$ as you have done. The reason the former is preferred is that it is monic, but aside from that, there are no real differences. This answer uses the convention you have used in your question, so the answers will differ from the other convention by a factor of $(-1)^n$.

As $p(t)$ is a polynomial of degree $n$, it has exactly $n$ zeroes counted with multiplicity over $\mathbb{C}$ by the Fundamental Theorem of Algebra. Therefore, $p(t) = a(t-\lambda_1)\dots(t-\lambda_n)$ for some $a$. Evaluating at $t = 0$ we see that $p(0) = a(-1)^n\lambda_1\dots\lambda_n = a(-1)^n\det A$. On the other hand, $p(0) = \det(A-0I) = \det A$. Equating the two expressions for $p(0)$, we see that $a = (-1)^n$. 
Now, by expanding the brackets, we have
\begin{align*}
p(t) &= (-1)^n(t - \lambda_1)\dots(t-\lambda_n)\\
&= (-1)^n\sum_{i=0}^n\left(\sum_{\lambda_{j_1} < \dots < \lambda_{j_i}}(-1)^i\lambda_{j_1}\dots\lambda_{j_i}\right)t^{n-i}\\
&= (-1)^n\sum_{i=0}^n\left((-1)^i\sum_{\lambda_{j_1} < \dots < \lambda_{j_i}}\lambda_{j_1}\dots\lambda_{j_i}\right)t^{n-i}\\
&=(-1)^n\sum_{i=0}^n(-1)^is_i(\lambda_1, \dots, \lambda_n)t^{n-i}
\end{align*}
where $s_i$ is the $i^{\text{th}}$ elementary symmetric polynomial. So the coefficient of $t^i$ is $(-1)^{i}s_{n-i}(\lambda_1, \dots, \lambda_n)$.
Let's check some specific cases:
If $\underline{i = 0}$, the coefficient of $t^0$ (i.e. the constant term) should be 
$$(-1)^0s_n(\lambda_1, \dots, \lambda_n) = \lambda_1\dots\lambda_n = \det(A),$$
which it is.
If $\underline{i = n}$, the coefficient of $t^n$ should be 
$$(-1)^ns_0(\lambda_1, \dots, \lambda_n) = (-1)^n1 = (-1)^n,$$
which it is.
Now for one you didn't know already. If $\underline{i = n-1}$, we see that 
$$(-1)^{n-1}s_1(\lambda_1, \dots, \lambda_n) = (-1)^{n-1}(\lambda_1 + \dots + \lambda_n) = (-1)^{n-1}\operatorname{trace}(A).$$
There is also a more highbrow way to interpret the coefficients of the characteristic polynomial: the coefficient of $t^i$ is $(-1)^i\operatorname{tr}(\bigwedge^{n-i}A)$ where $\bigwedge^{n-i}A$ is the linear map $\bigwedge^{n-i}\mathbb{R}^n \to \bigwedge^{n-i}\mathbb{R}^n$ given by $v_1\wedge\dots\wedge v_{n-i} \mapsto Av_1\wedge\dots\wedge Av_{n-i}$. 
