29 Functions of the matrix

Then

$$A^2=SD{S}^{-1}SD{S}^{-1}=S{D}^2{S}^{-1}$$

and so on so,

$$g(A)=a_m{A}^m+a_{m-1}{A}^{m-1}+\mathrm{\dots }+a_{0}I$$

$$=S(a_m{D}^m+a_{m-1}{D}^{m-1}+\mathrm{\dots }+a_{0}I)S^{-1}=Sg(D)S^{-1},$$

and we can easily check that $g(D)$ is as above.