Let \(A\) be an \(n \times n\) matrix with entries in the field \(F\) and let \(f_1, \ldots, f_n\) be the diagonal entries of the normal form of \(xI - A\). For which matrices \(A\) is \(f_1 \neq 1\)?
Solution
Exactly the scalar matrices have \(f_1\ne1\): \[ \boxed{f_1\ne1\iff A=cI\text{ for some }c\in F.} \]
Indeed, the first determinantal divisor is the gcd of the \(1\times1\) minors, i.e. of all entries of \(xI-A\), and in Smith normal form this gcd is \(f_1\) up to a unit [@HK71]. If some off-diagonal entry \(a_{ij}\) is nonzero, then \(xI-A\) contains the nonzero constant \(-a_{ij}\), a unit of \(F[x]\), so \(f_1=1\). Hence \(f_1\ne1\) forces \(A\) to be diagonal.
For a diagonal matrix \(A=\operatorname{diag}(a_1,\ldots,a_n)\), the nonzero entries of \(xI-A\) are \(x-a_1,\ldots,x-a_n\). Their gcd is nonconstant only if all these monic linear polynomials coincide, i.e. only if \(a_1=\cdots=a_n=c\). In that case \(A=cI\) and every nonzero entry of \(xI-A\) equals \(x-c\), so \(f_1=x-c\ne1\). This proves both directions.