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I have a set of 3 vectors, E = { [1, 1, -3], [2, -1, 3], [-6, 3, -9] } and I want to know if this previous set is a subset of F = { [1, 2, 3], [3, 2, 1], [1, 0, 1] }

$$ \left[ \begin{array}{ccc|ccc} 1 & 3 & 1 &1 &2 &-6\\ 2 & 2 & 0 &1 &-1 &3 \\ 3 & 1 & 1 &-3&3&-9\end{array} \right] Gauss-> \left[ \begin{array}{ccc|ccc} 1 & 3 & 1 &1 &2 &-6\\ 0 & -4 & -2 &-1 &-5 &15 \\ 0 & 0 & 2 &-4&7&-21\end{array} \right] $$

Now my answer : Yes, because there is no contradiction.

b) if F is a subset of E

$$ \left[ \begin{array}{ccc|ccc} 1 & 2 & -6 &1 &3 &1\\ 1 & -1 & 3 &2 &2 &0 \\ -3 & 3 & -9 &3&1&1\end{array} \right] Gauss-> \left[ \begin{array}{ccc|ccc} 1 & 2 & -6 &1 &3 &1\\ 0 & -3 & 9 &1 &-1 &-1 \\ 0 & 0 & 0 &9&7&1\end{array} \right] $$

Answer : No, because there is a contradiction.

Should I do something with the free variable (c3) here ?

Am I missing something here or the only thing I have to do is, gauss and check for a contradiction ?

Thanks

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Your original question asks whether you want the set $E$ to be a subset of $F$. I assume you mean is $E$ a subset of $\mathrm{span}(F)$? –  Christopher A. Wong Sep 20 '12 at 20:37
    
F is rank 3, therefore it spans $\mathbb R^3$, so any vector $\mathbf v$ in $\mathbb R^3$ will be in the span of F. E on the other hand is rank 2 (column 2 & 3 are linearly dependent) therefore, not all vectors in $\mathbb R^3$ are spanned by E. Part (b) shows that the columns of F are not spanned by E. –  Tpofofn Sep 20 '12 at 23:00
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1 Answer 1

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You're approaching this the right way:

With regards to the first part (assuming you want to show $E \subseteq \mathrm{span}(F)$), if you wish to determine whether or not e.g. $[1, 1, -3] \in \mathrm{span}(F)$, it's equivalent to determining whether or not the system of equations \[ \left[ \begin{array}{ccc} 1 & 3 & 1 \\ 2 & 2 & 0 \\ 3 & 1 & 1 \\ \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \end{array} \right] = \left[\begin{array}{c} 1 \\ 1 \\ -3 \\ \end{array} \right]\] has a solution, for $x_1,x_2,x_3 \in \mathbb{R}$. If it does have a solution, we have \[[1,1,-3] = x_1 [1,2,3] + x_2 [3,2,1] + x_3 [1,0,1].\] You seem to be doing it more efficiently by performing the Gaussian elimination on all three systems of equations at the same time.

However, having "no contradiction" is not a proof.

In your post, you've shown that the $3 \times 3$ matrix above has full rank (i.e. rank 3). So you can conclude that $\mathrm{span}(F)=\mathbb{R}^3 \supseteq E$. Alternatively, you could explicitly find the values of $x_1,x_2,x_3$ in each case by back substitution (thereby representing each element of $E$ as a linear combination of elements in $F$).

For the second part (assuming you want to show $F \subseteq \mathrm{span}(E)$), you have shown e.g. that the system of linear equations \[ \left[ \begin{array}{ccc} 1 & 2 & -6 \\ 1 & -1 & 3 \\ -3 & 3 & 9 \\ \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \end{array} \right] = \left[\begin{array}{c} 1 \\ 1 \\ 9 \\ \end{array} \right]\] is inconsistent. Therefore, $[1,1,9]$ cannot be written as a linear combination of the vectors $[1,2,3]$, $[3,2,1]$ and $[1,0,1]$. So $[1,1,9] \not\in \mathrm{span}(E)$.

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