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Given $$A= \left[ \begin{array}{cccc} 2 & 4 & 6 & 4 \\ 2 & 5 & 7 & 6 \\ 2 & 3 & 5 & 2 \end{array} \right] $$

I need to find the minimal spanning set of vectors for the row space of $A$.

My solution:

So I used Gaussian elimination on $A$ to get

$$ \left[ \begin{array}{cccc} 2 & 4 & 6 & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right] $$

$$ \left[ \begin{array}{cccc} 2 & 4 & 6 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$

$$ \left[ \begin{array}{cccc} 1 & 2 & 3 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$

My question is when you are trying to find the minimal spanning set of vectors for row space of A should I stop when I get it in the form of

$$ \left[ \begin{array}{cccc} 2 & 4 & 6 & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right] $$

Or carry on to get the final reduced matrix of

$$ \left[ \begin{array}{cccc} 1 & 2 & 3 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$

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  • $\begingroup$ You may consider using MathJax for better formatting. $\endgroup$ – Cave Johnson Sep 26 '16 at 14:42
  • $\begingroup$ sorry I was unable to use it for the matrices, if someone edits I will now use that format $\endgroup$ – user372523 Sep 26 '16 at 14:43
  • $\begingroup$ How did you go from one before the last matrix to the last one? You can't do that with elementary operations... $\endgroup$ – DonAntonio Sep 26 '16 at 14:45
  • $\begingroup$ i divided the first row by 2 $\endgroup$ – user372523 Sep 26 '16 at 14:48
  • $\begingroup$ or multiplying by a half $\endgroup$ – user372523 Sep 26 '16 at 14:52
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Find the span of $$ A= \left[ \begin{array}{cccc} 2 & 4 & 6 & 4 \\ 2 & 5 & 7 & 6 \\ 2 & 3 & 5 & 2 \end{array} \right]. $$ How precisely to approach this problem? There are a few ways. We will present an intuitive method, then a systematic method and discuss the different forms of the solution.


1. Opportunity-based method

Many academic exercises are simple enough to allow solution by inspection. In this case, $$ 2 \left( \text{row } 1 \right) - \text{row } 2 = \text{row } 3. $$ The first two rows are fundamental, so the span for the range of the row space is written as $$ \mathcal{R}\left( \mathbf{A}^{*} \right) = \text{span} \left\{ \, \left[ \begin{array}{c} 2 \\ 4 \\ 6 \\ 4 \end{array} \right], \left[ \begin{array}{c} 2 \\ 5 \\ 7 \\ 6 \end{array} \right] \, \right\} \tag{1} $$


2. Systematic method

The row reduction process. The following reduces the augmented matrix, but you could could by with only the naked matrix.

Approach 1

Column 1

$$ % operations \left[ \begin{array}{rcc} \frac{1}{2} & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{array} \right] % augmented matrix \left[ \begin{array}{cccc|ccc} 2 & 4 & 6 & 4 & 1 & 0 & 0 \\ 2 & 5 & 7 & 6 & 0 & 1 & 0 \\ 2 & 3 & 5 & 2 & 0 & 0 & 1 \\ \end{array} \right] % result = \left[ \begin{array}{crrr|rcc} \boxed{1} & 2 & 3 & 2 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 1 & 2 & -1 & 1 & 0 \\ 0 & -1 & -1 & -2 & -1 & 0 & 1 \\ \end{array} \right] $$

Column 2

$$ % operations \left[ \begin{array}{crc} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array} \right] % augmented matrix \left[ \begin{array}{crrr|rcc} \boxed{1} & 2 & 3 & 2 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 1 & 2 & -1 & 1 & 0 \\ 0 & -1 & -1 & -2 & -1 & 0 & 1 \\ \end{array} \right] % result = \left[ \begin{array}{crrr|rrc} \boxed{1} & 0 & 1 & -2 & \frac{5}{2} & -2 & 0 \\ 0 & \boxed{1} & 1 & 2 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -2 & 1 & 1 \\ \end{array} \right] $$

The boxed pivots identify the fundamental rows, and produce the same result in $(1)$.

Approach 2

Column 2

For this matrix, there are two fundamental rows. Any choice of the two will work. Here is a choice where we picks rows $1$ and $3$.

$$ % operations \left[ \begin{array}{ccr} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & -1 \\ \end{array} \right] % augmented matrix \left[ \begin{array}{crrr|rcc} \boxed{1} & 2 & 3 & 2 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 1 & 2 & -1 & 1 & 0 \\ 0 & -1 & -1 & -2 & -1 & 0 & 1 \\ \end{array} \right] % result = \left[ \begin{array}{cccr|rrr} \boxed{1} & 0 & 1 & -2 & -\frac{3}{2} & 0 & 2 \\ 0 & 0 & 0 & 0 & -2 & 1 & 1 \\ 0 & \boxed{1} & 1 & 2 & 1 & 0 & -1 \\ \end{array} \right] $$ In this instance the span is $$ \mathcal{R}\left( \mathbf{A}^{*} \right) = \text{span} \left\{ \, \left[ \begin{array}{c} 2 \\ 4 \\ 6 \\ 4 \end{array} \right], \left[ \begin{array}{c} 2 \\ 3 \\ 5 \\ 2 \end{array} \right] \, \right\} $$

Conclusion: your span could be written as any combination of two columns. $$ \mathcal{R}\left( \mathbf{A}^{*} \right) = % 1 & 2 \text{span} \left\{ \, \left[ \begin{array}{c} 2 \\ 4 \\ 6 \\ 4 \end{array} \right], \left[ \begin{array}{c} 2 \\ 5 \\ 7 \\ 6 \end{array} \right] \, \right\} = % 1 & 3 \text{span} \left\{ \, \left[ \begin{array}{c} 2 \\ 4 \\ 6 \\ 4 \end{array} \right], \left[ \begin{array}{c} 2 \\ 3 \\ 5 \\ 2 \end{array} \right] \, \right\} % 2 & 3 = \text{span} \left\{ \, \left[ \begin{array}{c} 2 \\ 5 \\ 7 \\ 6 \end{array} \right], \left[ \begin{array}{c} 2 \\ 3 \\ 5 \\ 2 \end{array} \right] \, \right\} \tag{2} $$

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