Cayley–Hamilton theorem

Definition

Let the Characteristic Polynomial of $A$ be $f_{\mathbf{A}}(\lambda )=\det (\mathbf{A}-\lambda \mathbf{I})={\sum }_{i=0}^n{a}_i{\lambda }^i$ Then, for the similar mapping $p_{\mathbf{A}}(\mathbf{A})={\sum }_{i=0}^n{a}_i{A}^i$, $p_{\mathbf{A}}(\mathbf{A})=0$ is satisfied

Proof

Let $\mathbf{B}=\mathrm{adj}(\lambda \mathbf{I}-\mathbf{A})$ Then, $\mathbf{B}(\lambda \mathbf{I}-\mathbf{A})=\det (\lambda \mathbf{I}-\mathbf{A})\mathbf{I}={\sum }_{i=0}^n{a}_i{\lambda }^i\mathbf{I}$ by the property of adjucate matrix

Since each element of $\mathbf{B}$ is $(n-1)$-order polynomial about $\lambda$ So we can express $\mathbf{B}={\lambda }^{n-1}{\mathbf{B}}_{n-1}+{\lambda }^{n-2}{\mathbf{B}}_{n-2}+\cdots +{\lambda }^1{\mathbf{B}}_1+{\mathbf{B}}_0$

Then, $\mathbf{B}(\lambda \mathbf{I}-\mathbf{A})={\lambda }^n{\mathbf{B}}_{n-1}+{\lambda }^{n-1}({\mathbf{B}}_{n-2}-{\mathbf{B}}_{n-1}\mathbf{A})+\cdots +\lambda ({\mathbf{B}}_0-{\mathbf{B}}_1\mathbf{A})-{\mathbf{B}}_0\mathbf{A}={\sum }_{i=0}^n{a}_i{\lambda }^i\mathbf{I}$ by using the two above expansions $\therefore {\mathbf{B}}_{n-1}=a_n\mathbf{I},({\mathbf{B}}_{n-2}-{\mathbf{B}}_{n-1}\mathbf{A})=a_{n-1}\mathbf{I},({\mathbf{B}}_0-{\mathbf{B}}_1\mathbf{A})=a_1\mathbf{I},-{\mathbf{B}}_0\mathbf{A}=a_0\mathbf{I}$

Therefore, $p_{\mathbf{A}}(\mathbf{A})={\mathbf{B}}_{n-1}{\mathbf{A}}^n+({\mathbf{B}}_{n-2}-{\mathbf{B}}_{n-1}\mathbf{A}){\mathbf{A}}^{n-1}+({\mathbf{B}}_0-{\mathbf{B}}_1\mathbf{A})\mathbf{A}+(-{\mathbf{B}}_0\mathbf{A})\mathbf{I}=0={\sum }_{i=0}^n{a}_i{\mathbf{A}}^i$

Examples

For $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$ $p(\lambda )=\det (\lambda I_2-A)=\det \left(\begin{array}{cc}\lambda -1& -2\\ -3& \lambda -4\end{array}\right)=(\lambda -1)(\lambda -4)-(-2)(-3)={\lambda }^2-5\lambda -2$ $p(A)=A^2-5A-2I_2=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$