Jordan form of operator $X \mapsto AXA$ Matrices $n \times n$ on complex field. Compute Jordan form of operator $X \mapsto AXA$:
$$
A = \begin{bmatrix} 
    0 & 1 & & \\
      & 0 & \ddots & \\
      &  & \ddots & 1 \\
         &    &    & 0 
    \end{bmatrix}
$$
A is nilpotent Jordan block
 A: Hint: The Jordan form of the map $T(X) = AXA$ can be deduced using (only) the following pieces of information:

*

*$T$ is a linear map on a space with dimension $n^2$

*$T^n = 0$

*More generally, $\operatorname{rank}(T^{k}) = (n-k)^2$, $k = 1,\dots,n$

Another approach: using the vectorization operator, we can conclude that the matrix of your transformation (relative to a certain basis of $\Bbb C^{n \times n}$) is $A^T \otimes A$, where $\otimes$ denotes the Kronecker product.  This matrix is "almost" in Jordan normal form.

To see that $\operatorname{rank}(T^{k}) = (n-k)^2$, $k = 1,\dots,n$ holds, it suffices to make the following observation.  The domain $\Bbb C^{n \times n}$ is spanned by elements of the form $uv^T$ with $u,v \in \Bbb C^n$.  Thus, the image of $T^k$ is spanned by elements of the form
$$
T^k(uv^T) = (A^k u)(v^T A^k) = (A^n u)((A^T)^kv)^T.
$$
Thus, the image of $T^k$ is spanned by the matrices $xy^T$ where $x,y$ are in the images of $A^k$ and $(A^T)^k$ respectively.  Because the image of $A^k$ and $(A^T)^k$ each have dimension $n-k$, we may conclude that the image of $T^k$ has dimension $(n-k)^2$.
Since $T$ is nilpotent, every Jordan block in the Jordan form is a block associated with $0$.  More specifically, we may use the above observation to conclude that $T$ has $1$ block of size $n$, and $2$ blocks of size $k$ for $k = 1,\dots,n-1$.
