How to tell if a set of vectors spans a space?

I want to know if the set {(1,1,1),(3,2,1),(1,1,0),(1,0,0)} spans R3. I know that if it spans R3, then for any x,y,z,R, there exist c1,c2,c3,c4 such that (x,y,z)=c1(1,1,1)+c2(3,2,1)+c3(1,1,0)+c4(1,0,0).

I’ve looked around the internet, but all the answers I found involve setting up a matrix and finding the determinant, and I can’t do that here because my matrix isn’t square. What am I missing here?

Answer

There are several things you can do. Here are four:

  1. You can set up a matrix and use Gaussian elimination to figure out the dimension of the space they span. They span R3 if and only if the rank of the matrix is 3. For example, you have
    (111321110100)(100321110111)(100021010011)(100010021011)(100010001001).
    (Sequence of operations: exchanged rows 1 and 4; subtracted first row from other rows to make 0s in first column; exchanged second and third rows; added multiples of the second row to third and fourth row to make 0s in the second column).

At this point, it is clear the rank of the matrix is 3, so the vectors span a subspace of dimension 3, hence they span R3.

  1. See if one of your vectors is a linear combination of the others. If so, you can drop it from the set and still get the same span; then you’ll have three vectors and you can use the methods you found on the web. For example, you might notice that (3,2,1)=(1,1,1)+(1,1,0)+(1,0,0); that means that
    span{(1,1,1), (3,2,1), (1,1,0), (1,0,0)}=span{(1,1,1), (1,1,0), (1,0,0)}.

  2. Determine if the vectors (1,0,0), (0,1,0), and (0,0,1) lie in the span (or any other set of three vectors that you already know span). In this case this is easy: (1,0,0) is in your set; (0,1,0)=(1,1,0)(1,0,0), so (0,1,0) is in the span; and (0,0,1)=(1,1,1)(1,1,0), so (0,0,1) is also in the span. Since the span contains the standard basis for R3, it contains all of R3 (and hence is equal to R3).

  3. Solve the system of equations
    α(111)+β(321)+γ(110)+δ(100)=(abc)
    for arbitrary a, b, and c. If there is always a solution, then the vectors span R3; if there is a choice of a,b,c for which the system is inconsistent, then the vectors do not span R3. You can use the same set of elementary row operations I used in 1, with the augmented matrix leaving the last column indicated as expressions of a, b, and c.

Attribution
Source : Link , Question Author : Javier , Answer Author : J. W. Tanner

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