Calculation of the power of the difference of two matrices (Proof of Hamilton-Cayley theorem)

Question

I am having trouble with a proof of the Hamilton-Cayley theorem (in the case the square matrix is diagonalisable).

$$ A=SDS^{-1} $$ $$ p_{A}(A)=(A-\lambda _{1}I)^{r_{1}}...(A-\lambda _{k}I)^{r_{k}} $$ $$ p_{A}(A)=S(D-\lambda _{1}I)^{r_{1}}...(D-\lambda _{k}I)^{r_{k}}S^{-1} $$

Where D is the diagonal matrix, the lambdas are the eigenvalues and p_A(A) is the characterstic polynomial of A.

In perticular I am having trouble understanding why when you replace A with SDS^-1 you can just take S and S^-1 out of the parentheses.

Answer 1

Since we have $A=SDS^{-1}$, we can easily take powers of $A$. For instance, \begin{equation} A^2=(SDS^{-1})(SDS^{-1}) = SDS^{-1}SDS^{-1} = SDIDS^{-1} = SD^2S^{-1}. \end{equation} Next, you should verify that $A^n=SD^nS^{-1}$, as well as that $A^n+A^m = S(D^n+D^m)S^{-1}$ for any $n,m\in\mathbb{N}$. Building up like this, eventually you can conclude that \begin{equation} p(A) = S~p(D)~S^{-1} \end{equation} for any polynomial $p$ (not just the characteristic polynomial).