Constant in front of the characteristic polynomial of a matrix In a text, I saw that the characteristic polynomial for an $n\times n$ matrix A with eigenvalues $e_{1}, \ldots e_{n}$ can be written
$$p(\lambda) = (\lambda - e_{1})(\lambda - e_{2}) \cdots (\lambda - e_{n}).$$
But shouldn't there be a constant in front of this polynomial? Like
$$p(\lambda) = K(\lambda - e_{1})(\lambda - e_{2}) \cdots (\lambda - e_{n}).$$
 A: The characheristic polynomial is of the form:
$$p(x)=
    \begin{vmatrix}
    x-a_{11} & . & . \\
    . & x-a_{22} & . \\
    . & . & x-a_{nn} \\
    \end{vmatrix}
$$
Now if we defold the determinant we see that the coefficient in the leading power of $x$ is $1$. That is why $K$ should be $1$.
A: In principle it could happen if we knew nothing about the characteristic polynomial's structure. 
However, it turns out that it will always be monic. One possible definiton for the determinant of a matrix is
$$
\det A = \sum_{\sigma \in \mathbb{S}_n}\operatorname{sgn}(\sigma) \cdot a_{1,\sigma(1)}\cdots a_{n,\sigma(n)}.
$$
Since the characteristic polynomial of a matrix $A$ is:
$$
\chi_A = \det(\lambda I - A) = \sum_{\sigma \in \mathbb{S}_n}\operatorname{sgn}(\sigma) \cdot (\lambda I -A)_{1,\sigma(1)}\cdots (\lambda I -A)_{n,\sigma(n)},
$$
this expression is a polynomial in $\lambda$, because it is a sum of producuts of either $\lambda - a_{ii}$ or a coefficient of $A$. The maximum exponent for $\lambda$ would occur, then, if $\lambda$ appears in every factor of a term, that is, in the term $(\lambda - a_{11})\cdots(\lambda - a_{nn})$, whose greater power of $\lambda$ is $\lambda^n$: recall that the degrees of a product is the sum of the degrees.
