# A subset of a linearly independent set of vectors is also linearly independent

Let $A = \{a_1, a_2, \dots, a_n \}$ be a set of vectors in $\mathbb{R^n}$ and let $B \subseteq A$. Prove that if $A$ is linearly independent, then $B$ must be linearly independent too.

The result is obvious intuitively, but I'm struggling with notation. Linearly independence means that $c_1a_1 + c_2a_2 + \dots c_na_n = 0 \iff c_1, c_2 \dots, c_n =0$.

Assume without loss of generality that $B = \{ a_1, a_2, \dots ,a_m\}$ with $m \leq n$, then $c_1a_1 + c_2a_2 + \dots + c_ma_m = 0 \iff c_1, c_2, \dots , c_m = 0$ as per the linear independence of $A$.

My question

• Is my usage of "without loss of generality" correct? Otherwise I have no idea how to denote the vectors of $B$.
• Is there a better way of making the last line of the proof clearer? I think the step "$c_1a_1 + c_2a_2 + \dots + c_ma_m = 0 \iff c_1, c_2, \dots , c_m = 0$" might be a bit too abrupt.
• Your proof is good. Commented Nov 29, 2016 at 18:36

\begin{align}&c_1a_1+\dots+c_ma_m=c_1a_1+\dots+c_ma_m+0a_{m+1}+\dots+0a_n=0\\ &\Leftrightarrow c_1=0,\dots,c_m=0,0=0\dots 0=0\\ &\Leftrightarrow c_1=0,\dots,c_m=0\end{align}
Loss of gerenality is ok. Order in a set is not relevant. Second part is OK, because $c_k$ can be $0$ for $k>m$.
I think the contrapositive gives a one liner. If $$B$$ is linearly dependent, then $$\sum_{i = 1}^m c_ia_i = 0$$ has a non-trivial solution. But then $$\left(\sum_{i = 1}^{n}c_ia_i + \sum_{i = m + 1}^n 0\cdot a_i\right) = 0$$ is a non-trivial solution so that $$A$$ is linearly dependent.