 Solution

The Cayley-Hamilton theorem states that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a proof, expand the adjugate matrix as $\mathrm{adj}(tI-A)={\sum }_{k=0}^{n-1}{B}_k{t}^k$. The adjugate identity givesComparing coefficients yields $-A{B}_0=c_{0}I$, $B_{k-1}-A{B}_k=c_{k}I$ for $1\le k\le n-1$, and $B_{n-1}=I$. Multiply the middle identities on the left by $A^k$, include the first identity, and add $A^n{B}_{n-1}=A^n$. The terms involving the $B_k$ cancel consecutively:Their right sides sum to $c_{0}I+{\sum }_{k=1}^{n-1}{c}_k{A}^k+A^n$. This proves the theorem without any substitution of noncommuting matrices into a scalar polynomial identity.