Let \(T\) be a linear operator on a finite-dimensional vector space over an algebraically closed field \(F\). Let \(f\) be a polynomial over \(F\). Prove that \(c\) is a characteristic value of \(f(T)\) if and only if \(c = f(t)\), where \(t\) is a characteristic value of \(T\).
Solution
1. Triangularize \(T\) and identify its characteristic values on the diagonal.
Because \(F\) is algebraically closed, the characteristic polynomial of \(T\) splits over \(F\). Hence \(T\) is triangularizable [@HK71]. Choose a basis in which \[ [T]=A= \begin{pmatrix} \lambda_1&*&\cdots&*\\ 0&\lambda_2&\cdots&*\\ \vdots&\ddots&\ddots&\vdots\\ 0&\cdots&0&\lambda_n \end{pmatrix}. \] The characteristic polynomial of a triangular matrix is \[ \prod_{i=1}^n(x-\lambda_i), \] so the characteristic values of \(T\) are exactly the numbers \(\lambda_1,\ldots,\lambda_n\), with repetitions according to algebraic multiplicity.
2. The diagonal entries of \(f(A)\) are \(f(\lambda_1),\ldots,f(\lambda_n)\).
Products and sums of upper-triangular matrices are upper triangular, and the diagonal of a product is the pointwise product of the diagonals. Hence, for every integer \(k\geq0\), the matrix \(A^k\) is upper triangular with diagonal \[ \lambda_1^k,\ldots,\lambda_n^k. \] If \[ f(x)=a_0+a_1x+\cdots+a_mx^m, \] then \[ f(A)=a_0I+a_1A+\cdots+a_mA^m \] is upper triangular with diagonal \[ f(\lambda_1),\ldots,f(\lambda_n). \] Moreover, matrix representation commutes with evaluating a polynomial in an operator, so \([f(T)]=f(A)\) in this basis.
3. Read off the characteristic values of \(f(T)\).
Since \(f(A)\) is triangular, its characteristic values are exactly its diagonal entries. By step 2 these are \[ f(\lambda_1),\ldots,f(\lambda_n). \] Combining this with step 1, a scalar \(c\) is a characteristic value of \(f(T)\) if and only if \[ c=f(\lambda_i) \] for some characteristic value \(\lambda_i\) of \(T\). Renaming \(t=\lambda_i\) gives precisely \[ c=f(t), \] as required.