Cayley-Hamilton with A=SDS^-1

Question

How would one prove the Cayley-Hamilton theorem for a case when A=SDS^−1, where A is diagonal? I have absolutely no idea.

Answer 1

Not sure what you mean. Take $$A=\begin{bmatrix}1&0\\0&2\end{bmatrix},\ \ \ S=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Then $p_A(t)=t^2-3t+2$. Also, $$ SAS^*=S=\begin{bmatrix}1&0\\0&0\end{bmatrix}, $$ so $$p_A(SAS^*)=\begin{bmatrix}0&0\\0&2\end{bmatrix}.$$

Answer 2

$\chi_A(t)=\det(tA-I)=\det(tSDS^{-1}-I)=\det(S(tD-I)S^{-1})=\det(tD-I)=\chi_D(t)$.

Therefore, $\chi_A(A)=\chi_D(SDS^{-1})=S\chi_D(D)S^{-1}=0.$

That is, you just need to prove the Cayley–Hamilton theorem for diagonal matrices, which is easy.

Indeed, $\chi_D(t)=(t-d_{11})\cdots(t-d_{nn})$. Since $De_i=d_{ii}e_i$, we have $\chi_D(D)e_i=0$ for all $i$ and so $\chi_D(D)=0$.