How to go from a recurrence relation to a final complexity

I have an algorithm, shown below, that I need to analyze. Because it's recursive in nature I set up a recurrence relation.

//Input: Adjacency matrix A[1..n, 1..n]) of an undirected graph G  
//Output: 1 (true) if G is complete and 0 (false) otherwise  
GraphComplete(A[1..n, 1..n]) {
  if ( n = 1 )
    return 1 //one-vertex graph is complete by definition  
  else  
    if not GraphComplete(A[0..n − 1, 0..n − 1]) 
      return 0  
    else 
      for ( j ← 1 to n − 1 ) do  
        if ( A[n, j] = 0 ) 
          return 0  
      end
      return 1
}

Here is what I believe is a valid and correct recurrence relation:

$\qquad \begin{align}
T(1) &= 0 \\
T(n) &= T(n-1) + n - 1 \quad \text{for } n \geq 2
\end{align}$

The "$n - 1$" is how many times the body of the for loop, specifically the "if A[n,j]=0" check, is executed.

The problem is, where do I go from here? How do I convert the above into something that actually shows what the resulting complexity is?

From what you wrote it seems that for all $k$ you have $T(k)=k-1+T(k-1)$ and $T(1)=0$. Therefore you can get it directly:

$$T(n)=(n-1)+(n-2)+\dots+1+0 = \sum_{k=1}^{n}{(k-1)}=\frac{n(n-1)}{2}$$

So $T(n)$ is in $Θ(n^2)$.

StackExchange Computer Science Q#1959, answer score: 10