= Solution
{c}
Put $w_1=v_1$ and $w_j=v_j-v_1$ for $j\geq2$. Since $v_1=w_1$ and $v_j=w_j+w_1$, the two families have the same <linear span>. Moreover,
$$
\sum_{j=1}^n a_jw_j
=\left(a_1-\sum_{j=2}^n a_j\right)v_1+\sum_{j=2}^n a_jv_j.
$$
If the $v_j$ are <linearly independent>, a vanishing expression forces $a_j=0$ for $j\geq2$ and then $a_1=0$. Conversely,
$$
\sum_{j=1}^n b_jv_j
=\left(\sum_{j=1}^n b_j\right)w_1+\sum_{j=2}^n b_jw_j,
$$
so <linear independence> of the $w_j$ forces all $b_j=0$. Thus both the <spanning set> condition and <linear independence> are equivalent for the two families, proving the desired equivalence of <bases>.
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