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Given that the linearly independent vectors $v_1$ and $v_2$ are linear combinations of the vectors $u_1$ and $u_2$ (also linearly independent), how can one prove that $span(\{v_1 ,v_2 \}) = span (\{ u_1, u_2\})$ ?

Very grateful for any help!

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  • $\begingroup$ You can't in general --- unless $u_1$ and $u_2$ are also linear combinations of $v_1$ and $v_2$.... $\endgroup$ – Lord Shark the Unknown Nov 8 '17 at 19:33
  • $\begingroup$ so they both generate two-dimensional subspaces, one of which is contained in the other. $\endgroup$ – Lord Shark the Unknown Nov 8 '17 at 19:44
  • $\begingroup$ yes, but is that enough to say for a proof? It makes sense intuitively, but is there any theorem or something that explicitly states that? Or is it simply common sense? $\endgroup$ – TaurusInIgnis Nov 8 '17 at 19:51
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You can't because $v_1$ and $v_2$ can be in the same subspace. Say for example in the usual basis given by $\{e_1,e_2\}$ on $\mathbb{R}^2$ we can have $v_1 = e_1 + e_2$ and $v_2 = 2e_1 + 2e_2$ and these two vectors do not span $\mathbb{R}^2$.

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  • $\begingroup$ Ah, I apologize for leaving out information. It is also given that u1 and u2 are linearly independent, as is the case with v1 and v2! $\endgroup$ – TaurusInIgnis Nov 8 '17 at 19:40

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