Coefficients of characteristic polynomial of a matrix For a given $n \times n$-matrix $A$, and $J\subseteq\{1,...,n\}$ let us denote by $A[J]$ its principal minor formed by the columns and rows with indices from $J$.
If the characteristic polynomial of $A$ is $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then why $$a_k=(-1)^{n-k}\sum_{|J|=n-k}A[J],$$
that is, why is each coefficient the sum of the appropriately sized principal minors of $A$?
 A: $\newcommand\sgn{\operatorname{sgn}}$
I learned of the following proof from @J_P's answer to what effectively is the same question. It arises from expanding the usual definition $\det A=\sum_{\sigma\in S_n}\sgn\sigma\prod_{1\le k\le n}A_{k,\sigma(k)}$, and deserves to be more well-known than it currently is.
Let $[n]:=\{1,\dots,n\}$, and write $\delta_{i,j}$ for the Kronecker delta, which is equal to $1$ if $i=j$, and is $0$ otherwise. Note that $\prod_{1\le k\le n}(a_k+b_k)=\sum_{C\subseteq[n]}\prod_{i\in C}a_i\prod_{j\in[n]-C}b_j$, since every term in the expansion on the left hand side will choose from each expression $(a_k+b_k)$ either $a_k$ or $b_k$, and so we may sum over all possible ways $C$ of choosing the $a_k$ terms.
We compute
\begin{align*}
\det(tI-A)
&=\sum_{\sigma\in S_n}\sgn\sigma\prod_{1\le k\le n}
(t\delta_{k,\sigma(k)}-A_{k,\sigma(k)})\\
&=\sum_{\sigma\in S_n}\sgn\sigma\sum_{C\subseteq[n]}
\prod_{i\in C}(-A_{i,\sigma(i)})\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}\\
&=\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_n}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}.
\end{align*}
For fixed $C\subseteq[n]$ and $\sigma\in S_n$, the last product $\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}$ vanishes unless $\sigma$ fixes the elements of $[n]-C$, in which case the product is just $t^{n-|C|}$. So we need only consider the contributions of the permutations of $C$ in our sum, by thinking of a permutation $\sigma\in S_n$ that fixes $[n]-C$ as a permutation in $S_C$. The sign of this permutation considered as an element of $S_C$ remains the same, as can be seen if we consider the sign as $(-1)^{T(\sigma)}$, where $T(\sigma)$ is the number of transpositions of $\sigma$. We thus have
\begin{align*}
\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_n}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}
&=\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_C}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}t^{n-|C|}\\
&=\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\sum_{\sigma\in S_C}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}.
\end{align*}
The term $\sum_{\sigma\in S_C}\sgn\sigma\prod_{i\in C}A_{i,\sigma(i)}$ is precisely the determinant of the principal submatrix $A_{C\times C}$, which is the $|C|\times|C|$ matrix with rows and columns indexed by $C$, and so
\begin{align*}
\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\sum_{\sigma\in S_C}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}
&=\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\det(A_{C\times C})\\
&=\sum_{0\le k\le n}\sum_{\substack{C\subseteq[n]\\|C|=k}}(-1)^kt^{n-k}
\det(A_{C\times C})\\
&=\sum_{0\le k\le n}t^{n-k}\left((-1)^k
\sum_{\substack{C\subseteq[n]\\|C|=k}}
\det(A_{C\times C})\right)\\
&=\sum_{0\le k\le n}t^k\left((-1)^{n-k}
\sum_{\substack{C\subseteq[n]\\|C|=n-k}}
\det(A_{C\times C})\right).
\end{align*}
A: Here's another way by using Taylor's theorem.
Consider $\det (xI+A)$ as a polynomial $p(x)$, from Taylor's theorem we have that:
$$
p(x)=\sum_{i=0}^n\frac{p^{(i)}(0)}{i!}x^i.
$$
Computing $p^{(i)}(0)$ will leads quikly to the conclusion.

How to compute $p^{(i)}(x)$ at $x=0$ ? Well, here's a trick:
For instance we compute $p'(0)$, go back to the determinant and replace the $x$ in the $k$th row by $x_k$, and using the total derivative. Then you'll find:
$$p'(0)=\sum_{|J|=n-1}A[J].$$
And using induction we can show in general that:
$$p^{(i)}(0)=i!\sum_{|J|=n-i}A[J]$$
A: Use the fact that $\begin{vmatrix} a & b+e \\
   c & d+f \end{vmatrix} = 
  \begin{vmatrix} a & b \\
   c & d \end{vmatrix} +   
  \begin{vmatrix} a & e \\
   c & f \end{vmatrix}
$
We can use this fact to separate out powers of $\lambda$. Following is an example for $2 \times 2$ matrix.
$$ 
\begin{vmatrix} a-\lambda & b \\
   c & d-\lambda \end{vmatrix} = 
  \begin{vmatrix} a & b \\
   c & d-\lambda \end{vmatrix} + 
  \begin{vmatrix} -\lambda & b \\
   0 & d-\lambda \end{vmatrix} = 
  \begin{vmatrix} a & b \\
   c & d \end{vmatrix} + %%
  \begin{vmatrix} a & 0 \\
   c & -\lambda \end{vmatrix} + %%
\begin{vmatrix} -\lambda & b \\
   0 & d \end{vmatrix} + 
  \begin{vmatrix} -\lambda & 0 \\
   0 & -\lambda \end{vmatrix}
$$
This decompose $det$ expression into sum of various powers of $\lambda$.
Now try it with a $3 \times 3$ matrix and then generalize it.
A: One way to see it: $A:V\to V$ induces the (again linear) maps $\wedge^k A:\wedge^k V\to \wedge^k V$. Your formula (restated in an invariant way, i.e. independently of basis) says that $$\det(xI-A)=x^n-x^{n-1}\operatorname{Tr}(A)+ x^{n-2}\operatorname{Tr}(\wedge^2 A)-\cdots(*)$$
 We can conjugate $A$ so that it becomes upper-triangular with diagonal elements $\lambda_i$ ($\lambda_i$'s are the roots of the char. polynomial). Now for upper triangular matrices the formula $(*)$ says that
$$(x-\lambda_1)\cdots(x-\lambda_n)=x^n-x^{n-1}(\sum\lambda_i)+x^{n-2}(\sum\lambda_i\lambda_j)-\cdots$$
which is certainly true, hence $(*)$ is true.
