# Suppose $S_1 =\{ u_1 , u_2 \}$ and $S_2 = \{ v_1 , v_2 \}$ are each independent sets of vectors in an n-dimensional vector space V.

Let us assume that every vector in $S_2$ is a linear combination of vectors in $S_1$.

Question: Does that mean that $S_1$ and $S_2$ are bases for the same subspace of $V$?

I know that the answer to this question is yes, subspaces spanned by both $S_1$ and $S_2$ are the same. Let's call them $W_1$ and $W_2$ respectively. How do we prove $W_1=W_2$? Equal subspaces when regarded as sets, must have the same elemnts. How can we show that EVERY vector in one subspace is also in the other subspace?

• Just show that the $v_i$ are a linear combination of the $u_i$ and vice versa. By assumption one direction is already done. – Krishan Bhalla Apr 28 '15 at 1:20

Let $\def\v#1{{\bf#1}}\v w\in W_2$. Then

• since $S_2$ is a basis for $W_2$ we have $\v w=\lambda\v v_1+\mu\v v_2$ for some scalars $\lambda,\mu$;
• from what you are given, $\v v_1=\alpha_1\v u_1+\alpha_2\v u_2$ and $\v v_2=\beta_1\v u_1+\beta_2\v u_2$;
• so $\v w=(\lambda\alpha_1+\mu\beta_1)\v u_1+(\lambda\alpha_2+\mu\beta_2)\v u_2$, and this is in $W_1$.

So $W_2\subseteq W_1$.

Since $W_2$ is a $2$-dimensional subspace of a $2$-dimensional space, the spaces are equal.

You don't need, At first, every vector of $S_2$ is linear combination of $S_1$, then $v_1,v_2$ are linear combinatios of $u_1,u_2$, so $span(S_2)\subset span(S_1)$

Then exists some $\alpha_1, \alpha_2, \beta_1, \beta_2$ such: $v_1=\alpha_1u_1+\alpha_2u_2$ and $v_2=\beta_1u_1+ \beta_2u_2$ Then solving the sistem for $u_1, u_2$, you can prove that their are linear comnbination of $v_1,v_2$ and for the first argument: $span(S_1)\subset span(S_2)$

$\therefore\;span(S_2)=span(S_1)$