Discussion
If V denotes the vector space of \(n \times n\) real skew symmetric matrices, then dim V =
Answer: D
The dimension is \(\dfrac{n(n - 1)}{2}\).
A skew symmetric matrix satisfies \(a_{ji} = -a_{ij}\). Taking \(i = j\) gives \(a_{ii} = -a_{ii}\), so every diagonal entry is 0. The entries above the diagonal can be chosen freely, and they fix those below. There are \(n(n - 1)/2\) positions above the diagonal, and the matrices \(E_{ij} - E_{ji}\) (\(i \lt j\)) form a basis.
\(n(n + 1)/2\) is the dimension of the symmetric matrices, which also keep the diagonal. \(n^{2} - n\) counts all off-diagonal positions without using the skew condition, and \(n - 1\) is the dimension of the diagonal matrices of trace zero.
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