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Let U = span{$u_1, u_2, u_3$},

where

$u_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} $, $u_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} $, $u_3 = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} $,

We are to find a basis {$w_1, w_2, w_3$} for U.

The first step that I have been taught to do this, is to have a matrix C with $u_1, u_2, u_3$ as rows. Why put them into rows?

(Did a brief check on similar questions, none of them answer my question. A brief Google search does bring up similar questions, but none of them answer why we put the {$u_1, u_2, u_3$} into rows.

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It is mainly for a technical reason: I guess you know Gauß's row reduction for solving linear equations. In a system of linear equations , it lets you detyermine the number of linearly independent equations.

For the span of a number of vectors, you should perform an analogous column reduction. It is simpler to transpose the matrix and perform row reduction.

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You put them as the rows of the matrix in order to determine whether or not they are all linearly independent. Try it --- when you put the vectors as the rows of a matrix and transform it into reduced echelon form, (2, 2, 2) will cancel because it is the combination of u1 + u2.

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  • $\begingroup$ Thanks for your comment. I'm well aware of row reducing the matrix. I'm just wondering why we put that into rows, and not row reduce the matrix with ${u_1, u_2, u_3}$ as columns instead. $\endgroup$ – MathsWanderer Sep 3 '15 at 21:28
  • $\begingroup$ I see. You could just as easily put the vectors as columns. It doesn't matter. If you decide to put the vectors as columns in order to find the basis, think about it as finding some non-zero vector r such that r1 * u1 + r2 * u2 + r3 * u3 = 0 where r1, r2, r3 are the components of vector r. (Sorry I'm terrible at formatting stuff). If you reduce that augmented matrix, you will end up with three equations in terms of r that describe the basis. $\endgroup$ – Deric Pang Sep 4 '15 at 17:39

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