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Let $S = \left(\left[\begin{array}{c} 1 \\ -1 \\ 3 \\ 2\end{array}\right], \left[\begin{array}{c} 2 \\ 1 \\ 1 \\ 3\end{array}\right], \left[\begin{array}{c} 1 \\ 5 \\ -7 \\ 0\end{array}\right], \left[\begin{array}{c} 4 \\ -1 \\ 7 \\ 7\end{array}\right]\right)$ and $V =$ span$(S)$. Find a basis of $V$.

I started by putting the vectors into the columns of a matrix. Then I row reduced the matrix and selected all the pivot columns as the vectors of the basis. However, when I checked the solutions, the vectors were placed into the matrix as rows instead of columns.

Does it matter if the vectors are placed as columns or rows?

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  • $\begingroup$ Row reducing the matrix of column vectors will also produce a basis, given that you take the pivot columns in the original matrix and not the row-reduced matrix. Both your method and the textbook's method will work. $\endgroup$ Commented Jun 30, 2022 at 19:43

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If you place them as column vectors, then you would have to apply column reduction or equivalently , row reduction on it's transpose. Row reduction ensures that you are taking linear combination of the rows. That is why the span remains unchanged.

So actually what you did was calculate a basis for the subspace spanned by $(1,2,1,4) , (-1,1,5,-1),(3,1,-7,7),(2,3,0,7)$ because when you applied elementary row operations on them, you took linear combinations of these vectors to get new vectors which have to lie in the span.

If you do it correctly, you should end up with a basis $\{(3,0,4,5) ,(0,3,-5,-1)\}$ .

A word of advice : When studying linear algebra, you will come across many topics which will involve algorithmic methods. For example, there is Gram-Schmidt process or Cramer's rule etc. But it will be wise to try and understand why these methods work. In otherwords, go through the proofs of these and try and understand them the best you can even if you are not that into abstraction and pure mathematics.

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