\[A \]
gives rise to a polynomial equation \[P( \lambda )=\lambda^n + c_{n-1}\lambda^{n-1}+... +c_1 \lambda +c_0=0 \]
(1)then
\[P(A)=A^n+ c_{n-1}A^{n-1}+... +c_1 A+c_0I=0\]
(2).It can be easily shown that these two equations are equivalent by diagonalizing the matrix
\[A \]
using the matrix \[P\]
of eigenvectors and the relationship \[D=P^{-1}AP=\]
where \[D\]
is the diagonal matrix of eigenvalues, then \[D^n=(P^{-1}AP)(P^{-1}AP)...(P^{-1}AP)=P^{-1}A^nP \]
so multiplying (2) on the left by\[P^{-1} \]
and on the right by \[P \]
we can write\[\begin{equation} \begin{aligned} P^{-1}(P( A))P &= P^{-1}A^nP + c_{n-1}P^{-1}A^{n-1}P+...+c_1 P^{-1}AP
+P^{-1}c_0P \\ &= D^n+c_{n-1}D^{n-1}+... +c_1D+c_0I =0 \end{aligned} \end{equation}\]
Now the equations (1) can be read off line by l;ine. (2) can be derived by the reverse process.