Constant in front of the characteristic polynomial of a matrix

Question

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

$$p(\lambda) = K(\lambda - e_{1})(\lambda - e_{2}) \cdots (\lambda - e_{n}).$$

Answer 1

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 $1$.

Answer 2

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.