= Solution
Taking the <matrix trace> of the adjugate expansion and using part (b) gives
$$
\sum_{i=0}^{n-1}\operatorname{Tr}(B_i)t^{n-1-i}
=\sum_{i=0}^{n-1}(n-i)c_it^{n-1-i}.
$$
Hence $\operatorname{Tr}(B_i)=(n-i)c_i$. Taking traces in $B_i=AB_{i-1}+c_iI$ gives, for $1\leq i<n$,
$$
c_i=-\frac1i\operatorname{Tr}(AB_{i-1}).
$$
For $i=n$, the same formula follows by taking the trace of $-AB_{n-1}=c_nI$. Iterating the recursion from part (a) also gives
$$
B_{i-1}=A^{i-1}+c_1A^{i-2}+\cdots+c_{i-1}I.
$$
Consequently the <Faddeev–LeVerrier algorithm> is
$$
\boxed{c_i=-\frac1i\left[\operatorname{Tr}(A^i)
+c_1\operatorname{Tr}(A^{i-1})+\cdots+c_{i-1}\operatorname{Tr}(A)\right],
\quad1\leq i\leq n}.
$$
Starting with $c_0=1$, this recursively expresses every $c_i$ using only $\operatorname{Tr}(A),\ldots,\operatorname{Tr}(A^i)$; these are the <Newton identities> for the <characteristic polynomial>.
Back to article page
