Dimension about space of matrices of order 3 over the field of the real. Consider the vector space of the matrices of order 3 over the field of the real $M_{3}\left(\Re\right)$ numbers. and let S be the subspace such that is spanned by the matrices of the form $AB-BA$. What is the dimension of S?
I was thinking in... the form of that matrix, and i know... If $X=AB−BA$ , then $trX=tr(AB−BA)=tr(AB)−tr(BA)=tr(AB)−tr(AB)=0$
well, i worked in an basis of that space... this:
$\left\{ \begin{pmatrix}0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix},\\ \begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
0 & 1 & 0
\end{pmatrix},\begin{pmatrix}1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -1
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 1
\end{pmatrix}\right\} $, 
it's fine the basis?

Let $X=AB-BA$ then $tr(X)=tr(AB-BA) = tr(AB) - tr(BA) = 0 $ we have the subspace of matrices with trace = 0, and I found the basis for the space:
$\left\{ \begin{pmatrix}0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix},\\ \begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 0 & 0\\
0 & 1 & 0
\end{pmatrix},\begin{pmatrix}1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -1
\end{pmatrix},\begin{pmatrix}0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{pmatrix},\right\} 
 $
then the dimension of the space is 8.
