Dimension of the space of symmetric traceless matrices A set of symmetric $n \times n$ matrices have $ \frac{1}{2}n^{2} - \frac{1}{2}n$ independent elements.  But how do you get to this result?  I understand that a general $n \times n$ matrix would have $n^{2}$  independent elements.  How exactly does the constraint of symmetry lead to the piece above?  Furthermore, how would I determine the number of elements for a symmetric traceless matrix?
An $n\times n$ matrix, $\mathbf T$, has, as you said, $n^2$ independent components, namely $[\mathbf{T}]_{ij}$, where each index can run from $1$ to $n$.
Symmetry, in this context, means that the matrix is equal to its transpose, $\mathbf{T}=\mathbf{T}^T$. What does this mean about the components? Let's see...

$[\mathbf{T}]_{ij}=[\mathbf{T}^T]_{ij}\\
   \hspace{0.85cm}=[\mathbf{T}]_{ji}$

That was simple enough! Symmetry implies that the $(i,j)$ component of the matrix equals the $(j,i)$ component of the matrix. What does this look like visually?

Neat - when you write out a symmetric matrix you'll notice that almost every component has a twin on the other side of the matrix. A helpful way to visualize a transpose is to flip the matrix across its diagonal and realize that symmetry means doing so does not change the matrix whatsoever. Long story short, when you get to counting independent components, be careful to not count both twins as if they were independent - only count pairs of twins.

Now, which elements DON'T have a twin? If you stare at the matrix, you'll soon figure out that it's the elements on the diagonal. That's 'cause $[\mathbf{T}]_{ii}=[\mathbf{T}]_{ii}$ trivially - symmetry did not impose any additional pairing up for these elements. Going back to our visual representation, if you flip a matrix across its diagonal, the diagonal elements stay put.
Phew, that was a lot, but now we're ready to count the number of independent components of a symmetric matrix. First the diagonal - there are $n$ independent components there. Now the pairs of twins. Removing the diagonal elements, we now have $n²-n$ components. If we twin them up, then we'll have $\frac{n²-n}{2}$ independent pairs. Then we just have to add the independent diagonal components back to get the total number of independent components:

$\frac{n²-n}{2}+n=\frac{n²-n+2n}{2}=\frac{n(n+1)}{2}$

Voila!
If you didn't like how we counted, then we can totally count them another way. Let's look at each row of the matrix and count how many independent components there are. Remember we're only counting the diagonal components and those on one side of the diagonal. The first row has $n$ independent components. The second row has $n-1$ independent components. We continue in this fashion till we get to the $n^{th}$ row which has but one independent component. Now we just have to add all these up:

$1+2+3+\cdots+(n-1)+n = \frac{n(n+1)}{2}$

If you're interested in how this sum was done, check out the really cool triangle numbers.

Now, about those traceless matrices. The trace is equal to the sum of all the diagonal components of the matrix. For a matrix to be traceless means that its trace is equal to zero. But that means that you can rewrite one of the diagonal components in terms of the others, so we have lost an independent component! Hence the number of independent components of a traceless symmetric matrix will be one less than the number of independent components of a symmetric matrix, namely

$\frac{n(n+1)}{2}-1=\frac{n^2+n-2}{2}=\frac{(n+2)(n-1)}{2}$

Hope that helps!
I'll$^1$ take an example from $SO(N)$: 

Consider $T^{ij}$ containing in total $N\times N = N^2$ indep. objects$^2$. But one can show that the symmetric, antisymmetric and the trace of $T^{ij}$ transform within themselves i.e. they don't mix with each other: 
$$\tag{def. trans. rule for each index}T^{ij}\rightarrow T^{'ij} = O^{il}O^{jm}T^{lm}$$
$$\tag{the symmetric part}S^{ij}\rightarrow S^{'ij} = O^{il}O^{jm}S^{lm}$$
$$\tag{the antisymmetric part}A^{ij}\rightarrow A^{'ij} = O^{il}O^{jm}A^{lm}$$
where $S^{ij} = 1/2(T^{ij}+T^{ji})$ and $A^{ij} = 1/2(T^{ij}-T^{ji})$. 

Finally the trace $T:=\delta^{ij}T^{lm}$ transforms as 
$$T\rightarrow T' = \delta^{ij}T^{'ij} = \delta^{ij}O^{il}O^{jm}T^{lm} = (O^T)^{li}\delta^{ij}O^{jm}T^{lm} = \delta^{lm}T^{lm} = T$$ where we used $O^TO=\mathbf{1}$. Thus we can define the symmetric traceless tensor from $T^{ij}$
$$Q^{ij} := S^{ij}-\frac{1}{N}\delta^{ij}T$$ which contains $\frac{1}{2}N(N+1)-1$ objects. 

In other words, one can split up the $N^2$ objects as

$$N\otimes N = [\frac{1}{2}N(N+1)-1]\oplus 1\oplus \frac{1}{2}N(N-1), $$
where the $1$ comes from the trace. 

To count the objects in the symmetric and asymmetric tensors just use (draw an $N$ by $N$ matrix and count the entries including the diagonal for the symmetric one, but not for the asymmetric one): 

$$\tag{for the symmetric part}\sum_{j=1}^{N} j = \frac{1}{2}N(N+1). $$
While for the asymmetric part we get 
$$\tag{for the antisymmetric part}\sum_{j=1}^{N} j - N= \frac{1}{2}N(N-1). $$
where the $N$ comes from the $N$ zeros on the diagonal. 

See also this PSE entry. 



$^1$Disclaimer: I'm not a mathematician so the below might contain errors and wrong terminology. Feel free to teach/correct me. 

$^2$With the number of objects contained in a tensor we mean the dimension of the representation. 

$^3$Much of this is taken from "QFT in a nutshell" by A. Zee.