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What is the dimension of the vector space of all symmetric matrices of order $n\times n$ $(n\geq 2)$ with real entries and trace equal to zero?

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Think of how you normally calculate the dimension of the space of symmetric matrices. The only difference comes in choosing elements on the diagonal: if I choose the first $n -1$ diagonal entries, then is there a choice for the last diagonal entry which will make the trace zero? How many choices for this last entry are there?

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You can choose arbitrary values for the upper triangular values, that gives $\frac{n(n-1)}{2}$ degrees of freedom, and you can choose arbitrary values for $n-1$ of the diagonal elements, so the dimension is $\frac{n(n-1)}{2}+(n-1)$.

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dear sir what if matrix have complex entries? –  srijan May 8 '12 at 16:37
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@srijan: Same answer. The choices have nothing to do with the field in question. –  copper.hat May 8 '12 at 16:44

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