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By solving the three vectors $(1,4,-3)$ , $(2,-7,5)$ and $(2,9,-7)$ ,
I got $x_1 = 0$ , $x_2 = 0$ and $x_3 = 0$.

I want to find the dimension of the subspace of $R^3$ spanned by these vectors.
How can I find it ?

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  • $\begingroup$ it spans a subspace, is linearly independent, so it is a basis. What's the dimension then? $\endgroup$ – user160738 Feb 10 '17 at 17:10
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Best to find the dimension of the subspace generated by the three vectors $(1,4,-3)$ , $(2,-7,5)$ and $(2,9,-7)$ is to consider the vectors as column vectors of a matrix and see the rank of the matrix.

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  • $\begingroup$ Thank You for your help :) $\endgroup$ – Student28 Feb 10 '17 at 19:08
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It looks like you've proof the vectors are linearity independent, so as they span the sub-space (by definition), they form a basis. Then the dimension is the cardinality of the basis, so 3

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Here $\alpha$$(1,4,-3)$+$\beta$$(2,-7,5)$+$\gamma$$(2,9,-7)$ = $(a,b,c)$ implies $\alpha$ = $a+6b+8c$ , $\beta$ = $\frac{a-b-c}{4}$ and $\gamma$ = $\frac{-a-11b-15c}{4}$, so that span{$(1,4,-3)$,$(2,-7,5)$,$(2,9,-7)$} = $\mathbb{R}^3$ and so the dimension is 3.

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