YX - XY = X for nilpotent matrix Let $X$ be a matrix over $\Bbb C$, I have to show that exists a matrix $Y$ s.t. $YX - XY = X$ iff $X$ is nilpotent.
What have I done? Given $Y$ exists, I have already shown that $tr(X^i)=0$ $\forall  i$ so $X$ is nilpotent.
I miss the converse.  
Thank you
 A: Let $X$ be given by
$$
X=\pmatrix{0&I_{n-1}\\0&0}.
$$
Every nilpotent $X$ can transformed by similarity to a block diagonal matrix with such diagonal blocks. Set $Y=diag(y_1\dots y_n)$.
Then
$$
YX-XY=\pmatrix{0&D\\0&0}
$$
with
$$
D=diag(y_1,y_2,\dots,y_{n-1}) - diag(y_2,\dots,y_n).
$$
In order to satisfy  $YX-XY=X$ it must hold $D=I$. Hence,
$$
Y=diag(n,n-1,\dots,1)
$$
does the trick.

My first idea was to construct $Y$ from powers of $X$. However, then $Y$ commutes with $X$, and $YX-XY=0$.
A: One way to go...(though abstract):  $X$ nilpotent means that you may decompose the space into nil-sequences. So let $e_{n-1}\mapsto e_{n-2} \mapsto \cdots \mapsto e_0 \mapsto 0$ be such a sequence under iteration by $X$. Define $Y e_k=-k e_k$, $0\leq k < n$. Then for $k\geq 1$: $$ (YX-XY)e_k=Y e_{k-1}+ k X e_k = -(k-1) e_{k-1} + k e_{k-1} = e_{k-1}=  X e_k$$
and $(YX -XY) e_0 = 0=Xe_0$. So $YX-XY=X$ on this subspace. Taking direct sums over nilsequences you get the stated claim. And no, I don't know (at least at present) how to write down this explicitly in terms of the matrix $X$ except in particular cases.
