Definition of minimal and characteristic polynomials I have defined the characteristic and minimal polynomial as follows, but have been told this is not strictly correct since det$(-I)$ is not necessarily 1, so my formulae don't match for $A=0$, how can I correct this?
Given an $n$-by-$n$ matrix $A$, the characteristic polynomial is defined to be the determinant $\text{det}(A-Ix)=|A-Ix|$, where $I$ is the identity matrix. The characteristic polynomial will be denoted by
\begin{equation}
\text{char}(x)=(x-x_1)^{M_1}(x-x_2)^{M_2}...(x-x_s)^{M_s}.\nonumber
\end{equation}
Also, we will denote the minimal polynomial, the polynomial of least degree such that $\psi(A)=\textbf{0}$, by 
\begin{equation}
\psi(x)=(x-x_1)^{m_1}(x-x_2)^{m_2}...(x-x_s)^{m_s}\nonumber
\end{equation}
where $m_{1}\le M_{1},m_{2}\le M_{2},...,m_{s}\le M_{s}$ and $\textbf{0}$ is the zero matrix.

There are two (nearly identical) ways to define the characteristic polynomial of a square $n\times n$ matrix $A$.  One can use either


*

*$\det(A-I x)$ or

*$\det(Ix-A)$


The two are equal when $n$ is even, and differ by a sign when $n$ is odd, so in all cases, they have the same roots.  The roots are the most important attribute of the characteristic polynomial, so it's not that important which definition you choose.  The first definition has the advantage that its constant term is always $\det(A)$, while the second is always monic (leading coefficient $1$).
With the minimal polynomial however, it is conventional to define it as the monic polynomial of smallest degree which is satisfied by $A$.
