Let's say set $S = \{(2,1,4), (1,-1,1), (3,2,5)\}$ and the vector space V is $R^3$. In this case, the number of independent vectors is equal to the dimension of the vector space. I know that one way to show that S is a spanning set for V is by showing that a linear combination of the vectors can equal any vector $(a,b,c)$. My professor told us to set up an augmented matrix and show that the rank of A is equal to the rank of A|b.
I would like to know if there is a less tedious method. If I should stick to this method, I would like to know why.
Edit
Let's say $S = \{(2,1,4), (1,-1,1), (3,2,5), (1,4,-2) \}$
How can I prove that a set spans a vector space if the number of independent vectors is larger than the dimension of the vector space? I won't be able to show that the set is a basis, and therefore also a spanning set. Is there a quick proof?
Edit 2 (for Future Reference)
The set S cannot have 4 independent vectors because the vectors are in $R^3$. If set S has more than 3 vectors it can still span $R^3$. That can be proven by finding a basis within the set.