Is a square matrix always similar to its transpose?
Solution
Yes. Let \(A\in M_n(F)\).
1. The polynomial matrices\[ tI-A \qquad\text{and}\qquad tI-A^t=(tI-A)^t \]have the same Smith normal form over the PID \(F[t]\).
If \[ U(tI-A)V=D \] is a Smith normal form with \(U,V\) unimodular, then transposing gives \[ V^t(tI-A)^tU^t=D^t=D. \] Thus the invariant factors are unchanged by transpose.
2. Therefore \(A\) and \(A^t\) have the same rational canonical form.
The invariant factors of \(tI-A\) are exactly the invariant factors of the \(F[t]\)-module associated to \(A\), and these determine the rational canonical form. By 1 the invariant factors for \(A\) and \(A^t\) agree.
3. Hence \(A\) is similar to \(A^t\).
Two matrices over a field are similar iff they have the same rational canonical form.