I have the following problem:
Find the rank of the operator on the vector space of real 3x3 matrices translating matrix X into AX-XA with $$A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{pmatrix}$$
I started with computing $AX-XA$ for a given $X=\begin{pmatrix} a & b & c\\ d & e & f\\ g & h & j \end{pmatrix}$
$$BX = \begin{pmatrix} 0 & 0 & 2c\\ 0 & 0 & 2f\\ -2g & -2h & 0 \end{pmatrix} \Rightarrow rank(BX) = 2 = dim(Im(B)) = rank(B)$$
But in the solution I have 4 as the answer. How is it possible? The maximal rank we can have here is 3. Or am I missing something?
The matrices $BX$ are spanned by FOUR vectors (corresponding to $c, f, g, h$) so the rank is equal to four. The fact that the elements of that vector space all have rank two is completely irrelevant.