Cayley-Hamilton theorem for diagonalizable matrices
Task number: 4466
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 \(p_\boldsymbol{A}(d_{jj})=0\).
