## Cayley-Hamilton theorem for diagonalizable matrices

### Task number: 4466

Show in a simple way that the Cayley-Hamilton theorem holds for diagonalizable matrices.
• #### Solution

Assume that $$\boldsymbol{A}=\boldsymbol{R}\boldsymbol{D}\boldsymbol{R}^{-1}$$, where $$\boldsymbol{D}$$ is diagonal. Every power $$\boldsymbol{D}^k$$ of a diagonal matrix $$\boldsymbol{D}$$ is diagonal and it has on the diagonal $$(\boldsymbol{D}^k)_{jj}=d_{jj}^k$$ i.e. powers of eigenvalues $$d_{jj}$$.

For $$p_\boldsymbol{A}(\boldsymbol{A})=\sum\limits_{i=0}^n b_i\boldsymbol{A}^i$$ it holds that $$p_\boldsymbol{A}(\boldsymbol{A})=p_\boldsymbol{A}(\boldsymbol{R}\boldsymbol{D}\boldsymbol{R}^{-1}) =\sum\limits_{i=0}^n b_i(\boldsymbol{R}\boldsymbol{D}\boldsymbol{R}^{-1})^i =\sum\limits_{i=0}^n b_i\boldsymbol{R}\boldsymbol{D}^i\boldsymbol{R}^{-1} =\boldsymbol{R}\Bigl(\sum\limits_{i=0}^n b_i\boldsymbol{D}^i\Bigr)\boldsymbol{R}^{-1} =\boldsymbol{R}\boldsymbol{0}\boldsymbol{R}^{-1} =\boldsymbol{0}$$, since the matrix $$\sum\limits_{i=0}^n b_i\boldsymbol{D}^i$$ has on the diagonal as its $$j$$-th entry $$p_\boldsymbol{A}(d_{jj})=0$$.