Let $n\in\mathbb{N}$ and $A\in M_n(\mathbb{F})$. Assume that the minimal polynomial of the square matrix $A$ is defined as the minimal polynomial of the linear operator $x\mapsto Ax$ for all $x\in\mathbb{F}^n$, on $\mathbb{F}^n$. Now, assume that the minimal polynomial of $A$ splits over $\mathbb{F}$. Show that $A$ and $A^T$ are similar.
Two matrices $A$ and $B$ similar if there exists an invertible matrix $S$ such that $B=SAS^{-1}$. Finding such an $S$ for $B=A^T$ will solve the problem.
Since the minimal polynomial of $A$ splits over $\mathbb{F}$, let us denote it by $$m_A(t) = (t-\lambda_1)^{m_1}…(t-\lambda_k)^{m_k}$$ for some $k\in\mathbb{N}$.
We know that $m_A(A)$ corresponds to the zero linear operator, i.e. $m_A(A)x=0$, $\forall x\in \mathbb{F}^n$. I quickly noticed that if matrices $P_1,P_2,…P_q$ commute, then $(P_1P_2…P_k)^T = P_1P_2…P_k$. It may be useful to replace each $P_i$ with $(A-\lambda_i)^{m_i}$?
I'm not sure how to proceed from here. I'd appreciate any hints!
Best Answer
You are overthinking it. If the minimal polynomial of $A$ splits over $\mathbb F$, $A$ and $A^T$ are similar to their Jordan forms over $\mathbb F$. However, as $\operatorname{rank}(\lambda I-A)^k=\operatorname{rank}(\lambda I-A^T)^k$ for each eigenvalue $\lambda$ of $A$ and each positive integer $k$, the two Jordan forms are the same (up to permutation of Jordan blocks). Hence $A$ and $A^T$ are similar.
Alternatively, that the minimal polynomial of $A$ splits over $\mathbb F$ implies that $A$ is similar to its Jordan form over $\mathbb F$. Since every Jordan block is similar to its own transpose via a reversal matrix (the matrix obtained by flipping the identity matrix left and right), $A$ is similar to $A^T$.