Let \(T\) be a linear operator on an \(n\)-dimensional space, and suppose that \(T\) has \(n\) distinct characteristic values. Prove that any linear operator which commutes with \(T\) is a polynomial in \(T\).
Solution
Let the distinct characteristic values of \(T\) be \[ \lambda_1,\ldots,\lambda_n. \]
1. The corresponding eigenvectors form a basis, and a commuting operator preserves each eigenline.
Choose a nonzero eigenvector \(v_i\) with \[ Tv_i=\lambda_i v_i. \] Eigenvectors belonging to distinct eigenvalues are linearly independent [@HK71]. Since there are \(n\) of them in an \(n\)-dimensional space, \(v_1,\ldots,v_n\) form a basis.
Let \(S\) be any linear operator satisfying \(ST=TS\). Then \[ T(Sv_i)=S(Tv_i)=S(\lambda_i v_i)=\lambda_i Sv_i. \] Thus \(Sv_i\) lies in the \(\lambda_i\)-eigenspace of \(T\). Because the \(n\) distinct eigenspaces already contribute \(n\) linearly independent eigenvectors, each eigenspace is one-dimensional. Hence for some scalar \(\mu_i\in F\), \[ Sv_i=\mu_i v_i. \]
2. Interpolate the scalars \(\mu_i\) by a polynomial.
Because the \(\lambda_i\) are distinct, the Lagrange polynomial \[ p(x)=\sum_{i=1}^n \mu_i\prod_{j\ne i}\frac{x-\lambda_j}{\lambda_i-\lambda_j} \] is well defined over \(F\) and satisfies \[ p(\lambda_i)=\mu_i \] for every \(i\).
3. The interpolating polynomial gives \(S=p(T)\).
For every basis vector \(v_i\), \[ p(T)v_i=p(\lambda_i)v_i=\mu_i v_i=Sv_i. \] Therefore \(p(T)\) and \(S\) agree on the basis \(v_1,\ldots,v_n\), so they are the same operator: \[ \boxed{S=p(T)}. \] Thus every operator commuting with \(T\) is a polynomial in \(T\).