Differing definitions of the Characteristic Equation According to the definition of an eigenvalue it satisfies the equation
$Ax=\lambda x$ where $A\in M_{n\times n}^{\mathbb{F}}$.
So that we could have either:
$(A-\lambda I)x=0$ or $(\lambda I-A)x=0$ 
such that the characteristic equation is either 
$det(A-\lambda I)=0$ or $det(\lambda I-A)=0$.
What practical differences result from the two possible definitions?
 A: There will be no difference in the roots of the characteristic equation. 
$
\left| \left(
\begin{matrix}
a_{11} - \lambda & a_{12} & ... & a_{1n} \\
a_{21} & a_{22} - \lambda & & \\
. \\
. \\
. \\
a_{n1} & ... & & a_{nn} - \lambda
\end{matrix}
\right) \right|
= 
(-1)^n \left| \left(
\begin{matrix}
\lambda - a_{11}  & - a_{12} & ... & - a_{1n} \\
- a_{21} & \lambda - a_{22} & & \\
. \\
. \\
. \\
-a_{n1} & ... & & \lambda - a_{nn}
\end{matrix}
\right) \right|
$
If $P(x)$ has a root $r$, then $-P(x)$ will have the same root.
A: Since $\det(cX) = c^n\det(X) $for an $n×n$ matrix, both are equivalent for even $n$.
A: The definition of the characteristic polynomial is$\det(\lambda I-A)$
and not $\det(A-\lambda I)$ because if $A$ is an $n\times n$ matrix
then the characteristic polynomial would of had a leading coefficient$-1$
and not $1$ for odd $n$.
In any case, there is not big difference here since if $\det(\lambda I-A)=P(\lambda)$
then $\det(A-\lambda I)=-P(\lambda)$ and in particular they have
the same roots.
Also note $(\lambda I-A)x=0\iff(A-\lambda I)x=0$ 
