Jordan canonical form of the transpose of a matrix

Question

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...

Answer 1

$A$ and $A^t$ have the same eigenvalues (why ?), hence the same Jordan canonical form. It follows that they are similar.

Answer 2

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.

Answer 3

Just because the Eigenvalues of two matrices are same does not guarantee similarity of Jordan form. Even if their characteristic, as well as minimal polynomial, are equal might not be enough to conclude similarity of their Jordan Form.

If you take a matrix S to be the matrix that has the standard basis vectors as its row, B$^{-1}$ = B = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix}

And, take, J = \begin{bmatrix} \lambda_{1} & 1 & & \\ & \ddots & \ddots & & \\ & & \lambda_{k} & 1 \\ & & & \lambda_{k} \end{bmatrix} ..

And, calculate B$^{-1}$J$^{T}$B then you will see that this is equal to J where J is the Jordan Form of the matrix A.

SO A and A$^{T}$ must be similar.