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I worked out that a matrix has to have the format \begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix} in order to be anti-symmetric. I think its basis might consist of these 3 matrices: \begin{bmatrix}0&1&0\\-1&0&0\\0&0&0\end{bmatrix} \begin{bmatrix}0&0&1\\0&0&0\\-1&0&0\end{bmatrix} \begin{bmatrix}0&0&0\\0&0&1\\0&-1&0\end{bmatrix}

Am I right? Or does its basis have 6 dimensions?

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  • $\begingroup$ Yes. No${}{}{}$. $\endgroup$ – Lord Shark the Unknown Sep 15 '17 at 15:00
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    $\begingroup$ Try to create the arbitrary matrix $$\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$$ from your three basis matrices. Does it work? If yes, you have a generating system - If no, you need more basis-vectors (matrices). If you have a generating system: What happens if you leave out one of the matrices? $\endgroup$ – P. Siehr Sep 15 '17 at 15:05
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$V$ decomposes into symmetric and anti-symmetric subspaces by the projections $A \mapsto (A \pm A^T)/2$ (which obviously sum to the identity map). One can apply these to the standard basis of $V$, composed of the matrices $E_{ij}$ which have a $1$ in the $(i,j)$th position and $0$s elsewhere, and obtain bases of the symmetric and antisymmetric subspaces.

Then $\{ E_{ij} + E_{ji} \mid i \geq j \}$ is a basis for the symmetric matrices (which then has dimension $4 \times 3/2=6$ since there are this many ways of choosing such pairs of $i$ and $j$), while $\{ E_{ij} - E_{ji} \mid i > j \}$ is a basis of the antisymmetric ones (which then have dimension $3 \times 2/2 = 3$ by the same idea).

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