Discussion
Let W be the vector space of real skew symmetric \(n \times n\) matrices. Then the dimension of W =
Answer: C
The dimension is \(\dfrac{n^{2} - n}{2}\).
A skew symmetric matrix has \(a_{ji} = -a_{ij}\), so every diagonal entry is 0 and the entries below the diagonal are fixed by those above. The free entries are the \(a_{ij}\) with \(i \lt j\), and there are \(n(n - 1)/2\) of them. The matrices \(E_{ij} - E_{ji}\) for \(i \lt j\) form a basis.
\(n^{2}\) is the dimension of all \(n \times n\) matrices, \(n^{2} - n\) counts all off-diagonal positions without using the skew symmetry, and \((n^{2} + n)/2\) is the dimension of the symmetric matrices.
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