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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.

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$\dim (\mathbb{R}^5)$ may be 5, but $\dim (W)$ is only 4. –  Yoni Rozenshein Sep 26 '12 at 0:00
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1 Answer

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.

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The problem was to find a subset of the original set of vectors that form a basis. While you could get the solution using an echelon form as you suggest, you need to keep track of row swaps. Note that the non-zero rows of an echelon form in general do not tell you that the same rows of the original matrix are independent. (It's easier to write the given vectors as columns, then row reduce to an echelon form. The columns of the echelon form that have pivots correspond to the columns of the original matrix that will form a basis.) –  David Mitra Sep 26 '12 at 1:25
    
David's got a point there, but it seems a rather easy task to keep track of rows swaps, particularly when compared to the extremely easy method proposed by user162...(+1) , of course. –  DonAntonio Sep 26 '12 at 9:58
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