Let \(S\) and \(T\) be square matrices over an algebraically closed field. Suppose \(S\) and \(T\) have the same Jordan canonical form. Prove that \(S\) and \(T\) are similar.
Solution
Let \(J\) denote their common Jordan canonical form.
1. There are invertible matrices \(P\) and \(Q\) such that\[ T=PJP^{-1}, \qquad S=QJQ^{-1}. \]
This is exactly what it means for \(J\) to be the Jordan canonical form of \(T\) and of \(S\).
2. Therefore \(S\) is similar to \(T\).
From the first equality, \[ J=P^{-1}TP. \] Substituting into the expression for \(S\) gives \[ S=QP^{-1}TPQ^{-1} =(QP^{-1})T(QP^{-1})^{-1}. \] Since \(QP^{-1}\) is invertible, this is a similarity transformation.