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Find a minimal spanning set of vectors for the row space of A.

Matrix $A$
$\begin{bmatrix} 2&4&6&4 \\2&5&7&6 \\2&3&5&2 \end{bmatrix}$

I believe to solve this problem I need to preform elementary row operations to the reduced row echelon for. So I have reduced my matrix to

$\begin{bmatrix} 1&2&3&2 \\0&1&1&2 \\0&0&0&0 \end{bmatrix}$

However I am not sure if I should multiply row 1 by 1/2 or not. If I did not I would get

$\begin{bmatrix} 2&4&6&4 \\0&1&1&2 \\0&0&0&0 \end{bmatrix}$

So here my solution would be. A minimal spanning set for row(A) is the vectors [2 4 6 4] and [0 1 1 2] or should it be [1 2 3 2] and [0 1 1 2]

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    $\begingroup$ Either is correct. $\endgroup$ – Kaj Hansen Sep 28 '16 at 23:23
  • $\begingroup$ so reduced row echelon form doesn't need leading 1's in each row? $\endgroup$ – jh123 Sep 28 '16 at 23:23
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    $\begingroup$ The definition of RREF specifies leading $1$'s in each row, but as far as a spanning set is concerned, notice, for example, that the pair of vectors $(1, 0)$ and $(0, 1)$ and the pair of vectors $(2, 0)$ and $(0, 2)$ will both span $\mathbb{R}^2$. $\endgroup$ – Kaj Hansen Sep 28 '16 at 23:25
  • $\begingroup$ okay so leading 1s is not a concern in spanning sets but you still need to get it in descending form on the diagonal $\endgroup$ – jh123 Sep 28 '16 at 23:26
  • $\begingroup$ Backing up to the title question, a minimal spanning set for the rowspace is a basis for the rowspace. A row echelon form gives one (as does the reduced row echelon form), but these are not the only ways a basis can be constructed. $\endgroup$ – hardmath Sep 29 '16 at 3:02

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