# Adjugate Matrix and Characteristic Polynomials in Wikipedia

If

$$p(t)~{\stackrel {\text{def}}{=}}~\det(t\mathbf {I} -\mathbf {A} )=\sum _{i=0}^{n}p_{i}t^{i}\in R[t],$$

is the characteristic polynomial of the matrix $n$-by-$n$ matrix $\mathbf A$ with coefficients in the ring $R$, then

$$\mathrm{adj}\,(s\mathbf{I}-\mathbf{A}) = \mathrm{\Delta}\! p(s,\mathbf{A}) ~, \tag 1$$

where

$$\mathrm {\Delta } \!p(s,t)~{\stackrel {\text{def}}{=}}~{\frac {p(s)-p(t)}{s-t}}=\sum _{j=0}^{n-1}\sum _{i=0}^{n-j-1}p_{i+j+1}s^{i}t^{j}\in R[s,t]$$

is the first divided difference of $p$, a symmetric polynomial of degree $n−1$.

I cannot make sense of equation $(1)$, since, for instance, $s-\mathbf A$ which would be in the denominator of $\mathrm{\Delta}\! p(s,\mathbf{A})$ is not well defined and even interpreting it as abuse of notation for $s\mathbf I-\mathbf A$ does not appear to help.

Am I just not understanding this correctly? Otherwise, is there a version of this that is actually correct?

The quotient $\frac{p(s)-p(t)}{s-t}$ has to be interpreted in the polynomial ring $R[s,t]$, where the numerator $p(s)-p(t)$ is in fact divisible by $s-t$, essentially because $$\frac{s^n -t^n}{s-t} = \sum_{i+j = n-1} s^i t^j.$$ This implies the given formula $$\Delta p(s,t) = \sum_{j=0}^{n-1} \sum_{i=0}^{n-j-1} p_{i+j+1} s^i t^j.$$ So in order to make sense of the expression $\Delta p(s,\mathbf A)$ you perform the division in $R[s,t]$ first and only then plug in the arguments $(s,\mathbf A)$, obtaining $$\Delta p(s,\mathbf A) = \sum_{j=0}^{n-1} \sum_{i=0}^{n-j-1} p_{i+j+1} s^i \mathbf A^j.$$