Cayley-Hamilton theorem and $\prod_{i=1}^nA-\lambda_iI$ ($\lambda_i$ is the $i$th eigenvalue)

Question

How does Cayley Hamilton theorem link to $\prod_{i=1}^nA-\lambda_iI$? All definitions of the theorem I can find say that $p(A)=0$ (the characteristic polynomial of $A$ is $0$). Thanks!

Answer 1

Over an algebraically closed field like $K=\mathbb C$, the characteristic polynomial of a square matrix $A\in K^{n\times n}$ factors as $$ p_A(x) = \prod_{i=1}^n (x-\lambda_i) = (x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n), $$ where the $\lambda_i$ are the eigenvalues of $A$ (counting multiplicity). Hence, $p_A(A)=0$ is equivalent to $$ \prod_{i=1}^n (A-\lambda_i I) = (A-\lambda_1 I)(A-\lambda_2 I)\cdots(A-\lambda_n I) = 0 \in K^{n\times n}. $$