Let \(A\) be an \(n \times n\) triangular matrix over the field \(F\). Prove that the characteristic values of \(A\) are the diagonal entries of \(A\), i.e., the scalars \(A_{ii}\).
Solution
Since \(A\) is triangular, so is \(xI-A\), and its diagonal entries are \[ x-A_{11},\ldots,x-A_{nn}. \] The determinant of a triangular matrix is the product of its diagonal entries. Therefore \[ \chi_A(x)=\det(xI-A)=\prod_{i=1}^n (x-A_{ii}). \] A scalar \(c\in F\) is a characteristic value exactly when \(\chi_A(c)=0\). Since \(F\) is a field, \[ \prod_{i=1}^n(c-A_{ii})=0 \] if and only if \(c=A_{ii}\) for at least one \(i\). Hence the characteristic values are exactly the diagonal entries of \(A\).