When $A$ is $n\times n$ matrix, prove that the coefficient of $\lambda^{n-1}$ in $p(\lambda)$ is $-{\operatorname{tr}}(A)$. 
When $A$ is $n\times n$ matrix, prove that the coefficient of  $\lambda^{n-1}$ in $p(\lambda)$ is $-{\operatorname{tr}}(A)$.

How can I prove this?
I can only think of a way to prove in triangular matrix:
$$(a_{11}-\lambda)(a_{22}-\lambda)\cdots(a_{nn}-\lambda)$$
 A: When doing the laplace expansion of $\det(\lambda I-A)$ along the first column, every minor except the one where you remove the first row and column has only $n-2$ entries with $\lambda$ in them, and so they make no contribution to the $\lambda^{n-1}$ term in the characteristic polynomial.  This observation lets you do an induction argument.  Explicitly, let $a_{11}$ be the top left entry of $A$, and let $A_{11}$ be the matrix obtained from $A$ by removing the first row and column.  Finally, let us use the notation $[\lambda^k]p(\lambda)$ to mean the coefficient of $\lambda^k$ in the polynomial $p(\lambda)$.  We then have
$$\begin{align}
[\lambda^{n-1}]\det(\lambda I-A)&=[\lambda^{n-1}](\lambda-a_{11})\det(\lambda I-A_{11}) \\
&=[\lambda^{n-1}]\lambda\det(\lambda I-A_{11})-[\lambda^{n-1}]a_{11}\det(\lambda I-A_{11}) \\
&=[\lambda^{n-2}]\det(\lambda I-A_{11})-a_{11}[\lambda^{n-1}]\det(\lambda I-A_{11}) \\
&=-\operatorname{tr}(A_{11})-a_{11} \quad \text{ (by the induction hypothesis)} \\
&=-\operatorname{tr}(A) 
\end{align}$$
A second algebraic approach is to use the formula
$$\det(A)=\sum_{\sigma\in S_n} (-1)^{\sigma}\prod a_{i\sigma(i)}.$$
If we use this to calculate the determinant of $\lambda I-A$, then the degree of the summand corresponding to a permutation $\sigma$ will be the number of fixed points of $\sigma$.  In particular, if we don't have at least $n-1$ fixed points, then the contribution will not have any $\lambda^{n-1}$ terms.  However, if a permutation on $n$ objects fixes at least $n-1$ of them, then it must fix all of them.  So the only term that contributes a $\lambda^{n-1}$ term is $\prod (\lambda-a_{ii})$.  It is straight forward to expand this out.
For a more geometric approach that works for matrices over algebraically closed fields, there is a result that every matrix is similar to an upper triangular matrix, which is a corollary to the fact that every square matrix has an eigenvector.  It is easy to compute the characteristic polynomial of a triangular matrix, and now we just need to use the fact that the characteristic polynomial and the trace are both preserved by conjugation.
For a more analytic approach, the collection of diagonaizable matrices is dense in the set of $n\times n$ matrices over $\mathbb C$, an argument similar to the above shows the result is true for diagonalizable matrices, and now you can use the fact that $\operatorname{tr}$ and the "characteristic polynomial" function (which maps matrices to polynomials) are both continuous functions.
A: Note that the trace is the sum of the eigenvalues, and then write the characteristic polynomial in terms of the eigenvalues (given that they are the roots of the characteristic polynomial). Now expand out the characteristic polynomial from this factorization (at least up to the $\lambda^{n-1}$ term).
A: Use Jordan canonical form $A=PDP^{-1}$, where $D$ is upper triagular matrix with the eigenvalues of $A$. Now Cayley-Hamilton theorem says if $p$ is the characteristic polynomial then $p(A)=p(P^{-1}DP)=P^{-1}p(D)P=0$.
So by Vieta's Formulas for polynomials the coefficient of $\lambda^{\deg(p)-1}$ is
$$\lambda_1+...+\lambda_n=-\text{tr}(D)=-\text{tr}(P^{-1}DP)=-\text{tr}(A)$$
Also you get $\det(A)=(-1)^n\lambda_1...\lambda_n$.
A: $$(a_{11}-\lambda)(a_{22}-\lambda)\cdots(a_{nn}-\lambda)=0 $$
$$ \implies (-1)^{n} (\lambda)^{n} + (-1)^{n-1} (a_{11}+a_{22}+\cdot \cdot \cdot + a_{nn}) \lambda^{n-1}+\cdot \cdot \cdot +(a_{11}a_{22}\cdot \cdot \cdot \cdot a_{nn}) =0 $$
$$\implies \lambda^{n} + (-1) (a_{11}+a_{22}+\cdot \cdot \cdot + a_{nn}) \lambda^{n-1}+\cdot \cdot \cdot +(-1)^{n} (a_{11}a_{22}\cdot \cdot \cdot \cdot a_{nn}) =0 $$
Roots of this $n$th degree equation are eigenvalues of $A$.
As sum of the roots of this $n$th degree equation is $$(-1)[(-1)(a_{11}+a_{22}+\cdot \cdot \cdot + a_{nn})] $$
(As, by the rule, for a $n$th degree equation, $$x^{n}+b_{1}x^{n-1}+b_{2}x^{n-2}+\cdot\cdot\cdot + b_{n}=0 $$, sum of all roots of this equation is $=-b_{1}$)
Hence , $tr(A) =$ sum of all roots =  $$(-1)[(-1)(a_{11}+a_{22}+\cdot \cdot \cdot + a_{nn})] $$
$$\implies (-1)(a_{11}+a_{22}+\cdot \cdot \cdot + a_{nn})= -tr(A) $$
Coefficient of $$\lambda^{n-1} =(-1)(a_{11}+a_{22}+\cdot \cdot \cdot + a_{nn}) = -tr(A) $$
