# Rank of transformation $Y=AX-XA$

Consider vector space $V$ consist all $n \times n$ matrix (real or complex).

What is the rank of the linear transformation $f(X)=AX-XA$ ($A\in V$)? ($A$ is a given matrix, which means we can have information about it)

I have tried to consider the basis of $V$ but it doesn't work.

EDITED:

$\operatorname{rank} f = n^2 - \operatorname{dim}N(f)$, which means we just need to find the demension of $N(f)$, which I think is much easier.

It turns out that it's not that easier.

EDIT 2: In the first answer it has been shown that the matrix version of the transformation has at least $n$ zero roots in the characteristic polynomial, however it's not true that these roots are accompanied with eigenvectors (Consider a nilpotent matrix of degree $n$ always have $n$ zero roots but may have only $1$ eigenvector). Therefore unless $A$ is diagonalisable, I think the problem is not solved yet.

• Considering $X$ is a $n$ x $1$ vector, and $A$ a $n$ x $n$ matrix, how can $XA$ be defined? – MathMan Jul 7 '14 at 10:57
• @VHP No, $X \in \mathbb K^{n\times n}$, so $f \colon V \to V$. – martini Jul 7 '14 at 10:58