Discussion
Let V denote the vector space of \(n \times n\) symmetric complex matrices, over \(\mathbf{R}\). Then dim V as a vector space over \(\mathbf{R}\) is :
Answer: C
The real dimension is \(n^{2} + n\).
A symmetric \(n \times n\) matrix is fixed by its entries on and above the diagonal: \(n\) diagonal entries and \(n(n - 1)/2\) above it, so \(n(n + 1)/2\) free entries. Over \(\mathbf{C}\) that is the dimension. Over \(\mathbf{R}\) each complex entry contributes two real parameters (its real and imaginary parts), so the dimension doubles:
\[ 2 \cdot \frac{n(n + 1)}{2} = n^{2} + n \]
\((n^{2} + n)/2\) is the complex dimension, \(n^{2}\) is the dimension of all real (or complex) matrices, and \(n^{2} - n\) has no such meaning here.
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