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I'm having trouble setting this up. I have these $3$ column vectors:

$\langle 1, 1, 2\rangle$

$\langle -7, -1, -8\rangle$

$\langle 3, 0, 3\rangle$

I need to find a linearly independent set of vectors that spans the same subspace of $\mathbb{R}^3$ as that spanned by these vectors.

When I do Gauss-Jordan, I end up with $x=0, y=0, z=0$. What I don't understand is how to find $2$ sets of vectors with this?

Any advice?

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up vote 3 down vote accepted

Set up the matrix $A$ whose rows are your vectors.

Use elementary row operations to bring it to reduced row-echelon form $B$.

Prove that the non-zero rows in $B$ are linearly independent and span the same subspace as is spanned by your original vectors.

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You don't have to do any of that, you can see that $\begin{bmatrix}-7\\-1\\-8\end{bmatrix}, \begin{bmatrix}3\\0\\3\end{bmatrix}$ will suffice as a linearly independent set of vectors that spans the same subspace of $\mathbb{R}^3$ as that spanned by the vectors since

"A set consisting of a pair of vectors is linearly dependent if and only if one of the vectors is a multiple of the other." (Source)

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But how do we know right away that the three original vectors are linearly dependent (or the span has dimension 2) without performing either a row reduction like described or some sort of similar calculation? (like writing one as a linear combination of the others). After we determine that the original 3 vectors form a subspace of dimension 2, then it is clear that this set of two vectors spans the same space. – Christian Jul 19 at 23:38
    
But if the problem states (as it does): find a linearly independent set of vectors that spans the same subspace of $\mathbb{R}^3$ as that spanned by these vectors we can assume that the three original vectors span has dimension 2 I would think. – user23 Jul 19 at 23:41
    
What if I gave, with the same question, the vectors: $(1, 0, 0), (2, 0, 0), (3, 0, 0)$? or $(1, 0, 0), (1, 1, 0), (1, 1, 1)$? In the first case they span a dimension 1 subspace, and the second they span a dimension 3 subspace. It might make the problem significantly easier in these cases (though, I could create a subspace that is harder to see that it is of dimension 3, like $(3, 4, -1), (7, 6, 2), (-3, 5, 1)$). As the question didn't stipulate that the vectors were linearly dependent to begin with, I think determining that is a necessary first step. – Christian Jul 19 at 23:50
    
Though, you are right, it probably wouldn't be a very good exercise if the vectors were linearly independent to begin with. – Christian Jul 19 at 23:57
    
I see @Christian – user23 Jul 20 at 0:02

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