CSIR JUNE 2011 PART B QUESTION 29 SOLUTION (Dimension of symmetric + trace zero matrices)

The dimension of the vector space of all symmetric matrices of order $n\times n$ ($n\ge 2$) with real entries and trace equal to zero is 1)$\frac{n^2-n}{2}-1$, 2)$\frac{n^2+n}{2}-1$, 3)$\frac{n^2-2n}{2}-1$ and 4)$\frac{n^2+2n}{2}-1$.
Solution: The set of all $n \times n$ real matrices $M_n(\Bbb R)$ has dimension $n^2$ since there are $n^2$ independent entries in a matrix to be filled. If a matrix is symmetric then we have $a_{ij} = a_{ji}$ for all $1 \le i,j \le n$. This shows that the entries above the diagonal are dependent and the only independent entries are the entries below the diagonal and on the diagonal. Now there are $1+2+\dots+(n-1) = \frac{(n-1)(n)}{2}$ many entries below the diagonal and $n$ entries in the diagonal. So the dimension of the symmetric matrices is $$\frac{(n-1)(n)}{2}+n = \frac{n^2+n}{2}.$$
But we have one more condition that trace is zero. This means that $a_{11}+a_{22}+\cdots+a_{nn}=0$. In particular $a_{11}=-(a_{22}+a_{33}+\cdots a_{nn})$. Hence the entry $a_{11}$ of the diagonal also dependent and only the remaining $n-1$ entries of the diagonal are independent. So the total independent entries are $$\frac{(n-1)(n)}{2}+(n-1) = \frac{n^2+n}{2}-1.$$ So option 2 is correct.
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