Discussion
Let V denote vector space of \(n \times n\) real matrices, having zero trace. Then dim V =
Answer: B
The dimension is \(n^{2} - 1\).
The trace is a linear map from the \(n^{2}\)-dimensional space of all \(n \times n\) real matrices onto \(\mathbf{R}\). It is onto (the matrix \(E_{11}\) has trace 1), so its image has dimension 1, and by rank-nullity its kernel, the matrices of trace zero, has dimension \(n^{2} - 1\).
Put differently, the single condition \(a_{11} + a_{22} + \ldots + a_{nn} = 0\) removes one degree of freedom from the \(n^{2}\) entries.
\(n^{2} - n\) would mean the whole diagonal is forced to zero, \(n(n - 1)/2\) is the dimension of the skew symmetric matrices, and \(n - 1\) is the dimension of the diagonal matrices with trace zero.
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