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.

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    $\begingroup$ You can't be sure that the leading coefficient of the characteristic polynomial will be positive, so the form you've written for $\operatorname{char}(x)$ is correct up to sign. Also, you should add the word 'monic' in your definition of minimal polynomial. Other than that, your definitions look good to me. Although, to write the minimal polynomial as you have requires Cayley-Hamilton. $\endgroup$ – Jared Apr 29 '13 at 15:54
  • $\begingroup$ So if I change my definition to be det$(Ix-A)=|Ix-A|$, then this will be correct? $\endgroup$ – hello123 Apr 29 '13 at 16:00
  • $\begingroup$ @julien: Thanks for catching that. I will delete and add another comment. $\endgroup$ – copper.hat Apr 29 '13 at 16:19
  • $\begingroup$ If c is a scalar, then $\det(cA)=c^n \det A$, since $\det$ is multilinear. In particular, $\det(a−Ix)=(−1)^n \det(Ix−A)$, where $n$ is the dimension of the space. $\endgroup$ – copper.hat Apr 29 '13 at 16:20

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

  1. $\det(A-I x)$ or
  2. $\det(Ix-A)$

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 $A$.

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