= Solution
{c}
The <vectors> are <linearly independent> if
$$
\sum_{j=1}^n a_jv_j=0,\quad a_j\in\mathbb R
\quad\Longrightarrow\quad a_1=\cdots=a_n=0.
$$
They <span> $V$ if every $v\in V$ is a <linear combination> $v=\sum_{j=1}^n a_jv_j$ with real <coefficients>. They form a <basis> of $V$ if they are <linearly independent> and <span> $V$. Equivalently, every $v\in V$ has exactly one such representation: existence is the <spanning set> condition, and subtracting two representations shows that uniqueness is precisely <linear independence>.
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