Multiply the invariant factors to recover the characteristic polynomial.Suggest a slogan
Let \(k\) be a field, let \(A\in\Mat_n(k)\), and let \(f_1\divides f_2\divides\cdots\divides f_n\) be the invariant factors of \(A\): the monic diagonal entries of the Smith normal form of \(xI-A\) over \(k[x]\), some of which may equal \(1\). Then \[ \det(xI-A)=\prod_{j=1}^nf_j(x). \]
The Smith normal form is \(P(xI-A)Q\) with \(P,Q\) invertible over \(k[x]\), so \(\det P\) and \(\det Q\) are nonzero constants and \(\det(xI-A)\) is a nonzero constant multiple of \(\prod_jf_j\). Both polynomials are monic, so they are equal.