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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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