dimension of a symmetric matrix with trace zero what will be the dimension of symmetric matrix of order $n\times n(n\geq2)$ with real entries and trace is equal to zero?
The answer is given as :$\frac{n^{2}+n}{2}-1$
can anyone explain how will get this answer?
according to me is should be:$\frac{n^{2}-n}{2}-1$
 A: You simply have to count all the numbers that you can put independantly in the matrix.
You can put all the numbers you want in one half or the other (but it will fix the numbers in the other half since the matrix is symetric), and all the elements in the diagonal, except one, because the trace being $0$ implies that the sum of the elements in the diagonal is $0$ (it's the definition of the trace), so the last number is fixed, its value being the opposite of the sum of all the others elements in the diagonal.
That gives $\frac{n^2+n}{2}-1$ independant values, so the dimension you want is $\frac{n^2+n}{2}-1$.
A: The dimension of the space of $n\times n$ symmetric matrices with diagonal equal to zero is $\frac{n^2-n}{2}$. Now, in your case the diagonal is not zero but the sum of its elements is zero, that means that you have $n-1$ elements which can vary. SO you get $\frac{n^2-n}{2}+(n-1)=\frac{n^2+n}{2}-1$ as expected.
A: Since each entry strictly below the diagonal is determined by a corresponding entry above the diagonal in a symmetric matrix, the dimension of the space of all such matrices is
$$
1 + 2 + \cdots + n = \frac{n^2 + n}{2}.
$$
The trace zero condition reduces the dimension by $1$.
A: If we define $T:$ your favorite subspace of square matrices$\to\mathbb R$ by $T(M)=$trace $(M)$,  then
$\dim\ker T+\dim im T= \dim$(your favorite subspace of square matrices).
So for zero trace subspace we have $\dim(YFSS)=\dim(YFSS)-1$.  
