Is my proof correct? Exercise about characteristic polynomials. This is an exercise from Hoffman's Linear Algebra: 
Let $A,B$ matrices of size $n$x$n$ on a given field $F$. We know that $AB$ and $BA$ have same eigenvalues. Do they have also same characteristic polynomial?
My answer: Yes, they do.
Proof. Let $p_1, p_2$ be the characteristic polynomial of $AB$ and $BA$, respectively. Notice that $p_1,p_2$ are monic polynomyals over $F$ of degree $n$. We know that there exist an extension $E$ of $F$, i.e., $F \leq E$ such that $p_1, p_2$ have $n$ roots in $E$. 
As we know that $AB$ and $BA$ have same eigenvalues, then we can conclude that $p_1,p_2$ have the same $n$ roots, and because they are monic, then $p_1 = p_2$.
Is my proof correct?
 A: Let $M=(m_{ij})$ be a $n\times n$ matrix, in a field $F$, and $0<r_1<r_2<\cdots <r_k\leq n$ and $0<s_1<s_2<\cdots <s_k\leq n$ be a sequence of numbers. One calls the following number
$$
M^{r_1, \cdots, r_k}_{s_1, \cdots, s_k} = \det\begin{pmatrix}
m_{r_1,s_1} & m_{r_1,s_2} & \cdots & m_{r_1,s_k}\\
m_{r_2,s_1} & m_{r_2,s_2} & \cdots & m_{r_2,s_k}\\
\vdots\\
m_{r_k,s_1} & m_{r_k,s_2} & \cdots & m_{r_k,s_k}
\end{pmatrix}
$$
is called a $k$-minor of $M$. In case $r_i=s_i$ once says $M_{r_1, \cdots, r_k}^{r_1, \cdots, r_k}$ is a principal $k$-minor. Now suppose we want to calculate the charcteristic polynomial of $M$. Write
$$
p_M(x)=\det(xI-M)=x^n-t_1(M)x^{n-1}+ t_2(M)x^{n-2}+\cdots + (-1)^nt_n(M)
$$
where $t_i(M)\in F$. It is easy to show that the coefficient $t_k(M)$, i.e. the coefficient of $x^{n-k}$, is the sum of all principal $k$-minors of $M$ (I leave it as an exercise). Formally
$$
t_k(M)=\sum_{0<r_1<r_2<\cdots <r_k\leq n} M_{r_1, \cdots, r_k}^{r_1, \cdots, r_k}
$$
Now the question is: Let $A,B$ be two $n\times n$ matrices. What is $(AB)_{r_1, \cdots, r_k}^{r_1, \cdots, r_k}$? This can be found using Cauchy-Binet formula. I will leave the details to you. For a statement of this formula and its proof see here. More concretely you want to prove:

Theorem: Let $A,B$ be two $n\times n$ matrices, $0<r_1<r_2<\cdots\leq r_k\leq n$ and 
  $0<s_1<s_2<\cdots\leq s_k\leq n$ two sequences of integers. Then
  $$
(AB)^{r_1, \cdots, r_k}_{s_1, \cdots, s_k}=\sum_{0<h_1<h_2<\cdots\leq h_k\leq n} A^{r_1, \cdots, r_k}_{h_1, \cdots, h_k}
B_{s_1, \cdots, s_k}^{h_1, \cdots, h_k}
$$

For brevity I will write $\sum_r$ instead of $\sum_{0<r_1<r_2<\cdots\leq r_k\leq n}$ from now on. Then
$$
\begin{aligned}
t_k(AB)=\sum_{r}(AB)^{r_1, \cdots, r_k}_{r_1, \cdots, r_k}=
\sum_{r}\sum_h A^{r_1, \cdots, r_k}_{h_1, \cdots, h_k}
B_{r_1, \cdots, r_k}^{h_1, \cdots, h_k}=\\
\sum_{h} \sum_r
B_{r_1, \cdots, r_k}^{h_1, \cdots, h_k}
A^{r_1, \cdots, r_k}_{h_1, \cdots, h_k}=\sum_h
(BA)^{h_1, \cdots, h_k}_{h_1, \cdots, h_k}=t_k(BA)
\end{aligned}
$$
which results in $p_{AB}(x)=p_{BA}(x)$. 
A: How do you know the roots of each polynomial come in the same multiplicities?  If you can pin these counts down, then your argument would work.
Another argument:
Suppose $A$ is invertible.  Then $BA=A^{-1}(AB)A$, so the characteristic polynomial of $BA$ is $\det(\lambda I-BA)=\det(A^{-1}(\lambda I-AB)A)=\det(\lambda I-AB)$.
If $A$ is not invertible, then there are ways to appeal to density in a suitable field extension.  There is also an argument (The characteristic polynomials of $AB$ and $BA$ by J.H. Williamson) which is universal, ignoring the issue of whether $A$ is invertible.  Consider the polynomial ring $\mathbb{Z}[\lambda][a_{ij},b_{ij}:1\leq i,j\leq n]$ in $2n^2+1$ variables, and consider the following equality (where $A$ and $B$ are the matrices $(a_{ij})_{ij}$ and $(b_{ij})_{ij}$, respectively):
$$\det(\lambda I-BA)\det(B)=\det(\lambda B-BAB)=\det(B)\det(\lambda I-AB)$$
Since the polynomial ring is an integral domain, it follows the polynomials $\det(\lambda I-BA)$ and $\det(\lambda I-AB)$ are equal.  The characteristic polynomials are the evaluations of these polynomials in $F[\lambda]$.
