On characteristic polynomial of $A+\lambda B$ Let $A,B$ be any $n\times n$ matrices over any field $F$ and $\lambda\in F$, non-zero.
Consider then the matrix $A+\lambda B$. I was trying to prove that the characteristic polynomial of this matrix is 
$$x^n + p_1(\lambda)x^{n-1} + \cdots + p_n(\lambda )$$
where $p_i(\lambda)$ are polynomials of degree $\leq i$ in $\lambda$.
I intuitively made it clear in the following way: assume simple case - $A$ being $0$; then think of this matrix $\lambda B$ with entries in $F$ as a matrix with entries in $\bar{F}$; we can conjugate it to an upper triangular matrix of the form 
$$\begin{bmatrix} \lambda b_{11} & * & \cdots & *\\ 0 & \lambda b_{22} &  & * \\  & & \ddots & \\ 0 & 0 & \cdots & \lambda b_{nn}\end{bmatrix}.$$
Then the characteristic polynomial of this matrix is 
$$(x-\lambda b_{11}) (x-\lambda b_{22}) \cdots (x-\lambda b_{nn})$$
which, after expansion, is easily seen to satisfy the desired expectation.
But, here, I did use of two facts: (1) Assumed $A=0$. (2) Looked the matrix over algebraic closure of field of consideration. 
In general, how can we argue without considering these specific cases or just staying in the field $F$?
 A: Write the expression as $$det(\cdots, A_i+\lambda B_i - xe_i, \cdots),$$ where the column written is the $i$-th and $A_i, B_i$ denote the $i$-columns of $A, B$ respectively. $e_i$ is the standard notation for the $i$-th vector of the canonical basis.
Since the expansion (using multilinearity) consists of terms with only one of the three terms above on each place, and its total degree is $\leq n$ (this is easy to see), the total degree is $\leq n$, and so the powers of $x$ accompanying $\lambda$ cannot have degree higher than $n-i.$
The expansion includes terms such as (for ease of notation we write a simple summand) $$\lambda^r(-x)^{n-r-s}det(A_1, \cdots, A_r, B_{r+1}, \cdots, B_{r+s}, e_{r+s+1}, \cdots, e_n).$$  In characteristic zero, you can obtain the terms in the expansion by taking suitable multiples of the partial derivative
$$\partial^a_\lambda \partial^b_x$$ of the polynomial, evaluated at $x=\lambda=0$ (use the Leibniz rule).
A: HINT
Use permutation-based definition of the determinant. 
Characteristic polynomial is $det(B+\lambda A -xI)$
Calculate it as plynomial of x and $\lambda$
if permutation has no elements in the diagonal - power  by x is 0, power by lambda is no more when $n$.
If permutation contains  k diagonal elements, see it debt as $P(\lambda)$ (not diagonal elements) * $P1(x, \lambda)$ (diagonal elements)
