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Let $V$ be a vector space over a field $F$, let $v_1, v_2, v_3 ∈ V$, and let
$u_1 = v_1, u_2 = 2v_1 + v_2$, and $u_3 = 3v_1 + 2v_2 + v_3$,
which are all elements of $V$. Prove the following:
• If $\{v_1, v_2, v_3\}$ is linearly independent, then $\{u_1, u_2, u_3\}$ is linearly independent.
• If $\{v_1, v_2, v_3\}$ spans $V$, then $\{u_1, u_2, u_3\}$ spans $V$

Note: I figured out the first part by plugging in the values for $\{u_1, u_2, u_3\}$ in a linear combination and equating it to zero, proving both the sets are linearly independent. Not sure how to prove that these vectors span the vector space $V$.

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Consider the equation $av_1+bv_2+cv_3=xu_1+yu_2+zu_3$ for span. $\implies$ $$x+2y+3z=a$$ $$y+2z=b$$ $$z=c$$ ($\because $ $v_1,v_2, v_3$ linearly independent) We get system of equations. using back substitution, we can prove that system has unique solution. Hence, proved. Both spans the same.

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Hint: If $v_1, v_2, v_3$ span $V$, then for an arbitrary vector $v$ in $V$ there exist scalars $c_1, c_2, c_3$ such that $$ v= c_1 v_1 + c_2 v_2 + c_3 v_3.\tag{*} $$ If $w_1, w_2, w_3$ also span $V$, then there exist scalars $d_1, d_2, d_3$ such that $$ v= d_1 w_1 + d_2 w_2 + d_3 w_3. $$

Can you substitute the given formula for the $w_i$’s in terms of the $v_i$’s into equation (*) in order to find the $d_i$’s (in terms of the $c_i$’s)?

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