Finding subset of vectors which form a basis Let $W = \{x \in \Bbb{R}^5 | \sum_{i=1}^5 x_i=0\}$. The following vectors span W. Find a subset of the following vectors which forms a basis for W.
$u_1 = (2, -3, 4, -5, 2)$
$u_2 = (-6, 9, -12, 15, -6)$
$u_3 = (3, -2, 7, -9, 1)$
$u_4 = (2, -8, 2, -2, 6)$
$u_5 = (-1, 1, 2, 1, -3)$
$u_6 = (0, -3, -18, 9, 12)$
$u_7 = (1, 0, -2, 3, -2)$
$u_8 = (2, -1, 1, -9, 7)$  
I understand what is required to solve this problem. $Dim(\Bbb{R}^5) = 5$, so I need to throw away 3 vectors that are linearly dependent. I am wondering if there is a simple way to solve this problem, instead of checking if each vector depends on the others.
 A: There's a way using matrices. Write each vector as a row in a matrix. You'll get a matrix with 5 columns and 8 lines. Then you have to reduce the matrix. Some lines will become filled with zeroes, but that's because that line was a linear combination of the others (so that vector was linear combination of the others).
Once the matrix is in the reduced form you'll have lines filled with zeroes and lines that are not filled with zero. Then you turn back the lines that are not null to the notation of vectors, and those vectors will form a basis.
Can you try to justify this steps? It's easy to show that it works, it's just needed to think about the procedure used in reducing a matrix.
I hope this helps you a little.
A: By observation u2 is a multiple of u1.
I would suggest pick u1, then consider whether u3 is a linear combination of u1 (since u2 is clearly a multiple), then see if u4 is a linear combination of {u1,u3}, and so on. You only need to find 4 vectors to span W (dim(W)=4), and since you have {u1,u3} (by observation) you only need to find two more vectors, so try u4 and u5
