Let \(M\) be a square matrix over a field \(K\). Use a suitable canonical form to show that \(M\) is similar to its transpose \(M^T\).
Solution
We use the Rational Canonical Form (Frobenius normal form), which exists for any square matrix over an arbitrary field \(K\).
Over the field \(K\), \(M\) is similar to a direct sum of companion matrices of its invariant factors: \[ M \sim C(p_1) \oplus C(p_2) \oplus \cdots \oplus C(p_k), \] where each \(C(p_i)\) is the companion matrix of a monic polynomial \(p_i(x) = x^d + a_{d-1}x^{d-1} + \cdots + a_1 x + a_0 \in K[x]\).
Since transposition distributes over direct sums and preserves similarity: \[ M \sim M^T \iff C(p_i) \sim C(p_i)^T \text{ for each } i=1, \ldots, k. \] Thus, it suffices to prove that any companion matrix \(C = C(p)\) is similar to its transpose \(C^T\).
Let \(C\) be the \(d \times d\) companion matrix of \(p(x)\): \[ C = \begin{pmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{d-1} \end{pmatrix}. \] The characteristic and minimal polynomials of both \(C\) and \(C^T\) are equal to \(p(x)\). Since \(C^T\) has minimal polynomial \(p(x)\) of degree equal to the dimension \(d\), \(C^T\) is also cyclic, and its single invariant factor is \(p(x)\).
By the uniqueness of the Rational Canonical Form, any two matrices over \(K\) with the same invariant factors are similar over \(K\). Since \(C\) and \(C^T\) have the identical list of invariant factors (namely, the single polynomial \(p(x)\)), \(C\) is similar to \(C^T\) over \(K\).
Taking the direct sum over all blocks gives \(M \sim M^T\) over \(K\).