Basis for Skew Symmetric Matrix I'm trying to find a basis for the kernel for the following mapping:  Considering the linear transformation T: $M_{33} \rightarrow M_{33}  $ defined by $T(A) = .5(A + A^T)$. I know that this is basically asking for the basis under the condition that $T(A)=0$ which means that $A+A^T=0$ so $A^T = -A$.  I found that matrices that fit this condition are Skew Symmetric Matrices.  However, I'm not sure how to find the basis for the kernel of these matrices. 

Let $a_{ij}$ denote the entries of $A$.  If $A \in \ker T$, then all of the entries of $T(A)$ are zero.  In other words,
$$
a_{ij} + a_{ji} = 0.
$$
This forces diagonal entries to vanish:
$$
a_{ii} = 0.
$$
Define the matrix unit $E_{ij}$ to be the $3 \times 3$ matrix, all of whose entries are $0$ except for the $(i,j)$ entry, which is $1$.  These nine matrices form a basis for $M_{3,3}$, the space of all $3 \times 3$ matrices.  
Now, we can build a basis $\{ B_{12}, B_{13}, B_{23} \}$ for the space of skew symmetric matrices out of the matrix units:
\begin{align}
B_{12} = E_{12} - E_{21} &= \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\!, \\[2pt]
B_{13} = E_{13} - E_{31} &= \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}\!, \\[2pt]
B_{23} = E_{23} - E_{32} &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}\!.
\end{align}
An arbitrary skew symmetric matrix decomposes as
$$
\begin{pmatrix} 0 & a_{12} & a_{13} \\ -a_{12} & 0 & a_{23} \\ -a_{13} & -a_{23} & 0 \end{pmatrix}
= a_{12} B_{12} + a_{13} B_{13} + a_{23} B_{23}\!,
$$
showing that the set $\{ B_{12}, B_{13}, B_{23} \}$ spans.  It's pretty clear that these three are linearly independent as well: if we set the arbitrary linear combination to zero on the right, then each entry of the matrix is $0$, so $a_{12} = a_{13} = a_{23} = 0$ which is the trivial combination.  In other words, the decomposition of any skew symmetric matrix is unique.
