# Proving span of given vectors equals the span of their linear combinations

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!

• You can't in general --- unless $u_1$ and $u_2$ are also linear combinations of $v_1$ and $v_2$.... – Angina Seng Nov 8 '17 at 19:33
• so they both generate two-dimensional subspaces, one of which is contained in the other. – Angina Seng Nov 8 '17 at 19:44
• 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? – TaurusInIgnis Nov 8 '17 at 19:51

## 1 Answer

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 $\mathbb{R}^2$.

• 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! – TaurusInIgnis Nov 8 '17 at 19:40