# Jordan form of operator $X \mapsto AXA$ [closed]

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

• do you mean $X \mapsto AXA$? Also, what have you tried? Jul 22, 2019 at 1:37

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 $$k = 1,\dots,n-1$$.