Dimension of $W$ Where $W$ is the subspace of matrices with trace=0 
Let $V$ be the vector space of $n*n$ symmetric matrices.And let $W$ be the
  subspace of $V$ consisting of those Matrices $A$ such that $tr(A)=0$.
  What is the dimension of $W$?

What I have noted is that the dimension of $V$ is $\frac{n^2-n}{2}+n$ (By considering the possible arrangements).
And when it comes to the subspace $W$,
To get a the basis, First we will need all the basis elements that create the non diagonal matrices. Thus the basis should have $\frac{n^2-n}{2}$ elements (Note that the vector space is symmetric).
Then in addition to that we will need a relationship with the diagonal elements.
Say for example the vector space of $3*3$ matrices,
$$B=\begin{pmatrix}
b_1 &  & \\ 
 &  b_2& \\ 
 &  & b_3
\end{pmatrix}$$
if $B\in W$ then $b_1=-b_2-b_3$
So the basis of $W$ should have,
 $\begin{pmatrix}
-1 & 0 &0 \\ 
0 &  1&0 \\ 
 0&  0& 0
\end{pmatrix}$, $\begin{pmatrix}
0 & 0 &0 \\ 
0 &  -1&0 \\ 
 0&  0& 1
\end{pmatrix}$.
And even for a general $n$ dimension set also this seems correct. But I would like a feedback of an explicit proof
 A: Note that $\operatorname{tr}: S_n \to \mathbb{C}$ is linear and non zero. Hence $\dim \ker \operatorname{tr} = \dim S_n -1$.
A: The dimension of the space of symmetric matrices is $\frac{1}{2}n(n+1)$, which are the the triangular numbers. You only have to give the entries on the main diagonal, and then the top-right triangle. That's the $n$-th triangular number.
If you prefer: $1+2+3+\cdots+n = \frac{1}{2}n(n+1)$.
Once you've fixed those, the rest of the entries follow by symmetry.
If you want the trace to be zero then you impose once more condition, so the dimension is
$$\frac{1}{2}n(n+1)-1 = \frac{1}{2}(n-1)(n+2)$$
A: Let $S_n$ be the space of symmetric matrices, and let $T_n$ be the space of matrices of trace zero.
We know that $\dim(S_n) = \frac{n^2-n}2+n$ and $\dim(T_n) = n^2-1$. Notice that $S_n + T_n = M_n(\mathbb{R})$. This is because for every matrix $A \in M_n(\mathbb{R})$ holds
$$A = \frac12(A + A^T) + \frac12(A - A^T)$$
where the first summand is symmetric and the second has trace zero.
Therefore we have
\begin{align}\dim W &= \dim (S_n \cap T_n) \\
&= \dim(S_n) + \dim(T_n) - \dim(S_n + T_n) \\
&= \left(\frac{n^2-n}2+n\right)+(n^2-1)-n^2 \\
&= \frac{n^2-n}2+n-1\\
&= \frac{(n+2)(n-1)}2
\end{align}
