Jordan canonical form of the transpose of a matrix Let $A \in M_{n \times n}(\mathbb C)$. Prove that $A$ and $A^\text{T}$ have the same Jordan canonical form, and conclude that $A$ and $A^\text{T}$ are similar.

Intuitively, I know that I need to flip each cycle of generalised eigenvectors of $A^\text{T}$, but how to prove it?

Moreover, I've tried Mathematical Induction but was stuck...
Transposed matrices have the same characteristic equation and minimal polynomial. Furthermore, they have the same generalised eigenspace. With all of this in mind, we can say that $A$ and $A ^ \text{T}$ have the same Jordan chains, and hence Jordan basis. So $A$ and $A ^ \text{T}$ have the same Jordan canonical form.

Finally, similar matrices have the same Jordan matrix because if $P, Q \in \text{GL}(n, \mathbb{C})$ and $J$ is the Jordan canonical form, then we can write
$$P^{-1}AP = J = Q^{-1}A^\text{T}Q$$
i.e.
$$QP^{-1}APQ^{-1} = A^\text{T}.$$
But $Q, P ^{-1}$ and $P, Q^{-1}$ are invertible, so $QP^{-1}$ and $PQ^{-1}$ are also invertible. So write $R = QP^{-1}$. Then we know that $R^{-1} = (QP^{-1})^{-1} = (P^{-1})^{-1}Q^{-1} = PQ^{-1}$. So

$$RAR^{-1} = A^\text{T}$$

and we can conclude that $A$ and $A^\text{T}$ are similar.
$A$ and $A^t$ have the same eigenvalues (why ?), hence the same Jordan canonical form.
It follows that they are similar.