singleton null vector set linearly dependent, but other singletons are linearly independent set Why the set $\{\theta_v\}$ where $\theta_v$ is the null vector of a vector space is a dependent set intuitively (what is the source of dependence) and the singleton vector set which are non-null are independent sets ? (btw, I know how to show it mathematically, but not understanding the intuition behind this).
Does it really make sense to think of independence of a singleton set ?  
 A: The intuition is the following: the null vector only spans a zero-dimensional space, whereas any other vector spans a one-dimensional space. This is captured by the following thought:
A set of vectors $\{ \bar v_1, \bar v_2, ..., \bar v_n\}$ is linearly independent iff $span(\{ \bar v_1, ..., \bar v_n\})$ is not spanned by a proper subset of $\{ \bar v_1, \bar v_2, ..., \bar v_n\}$. Now, the space spanned by $\{ \bar o\}$ is already spanned by a proper subset of $\{ \bar o\}$ namely $\emptyset$. For a non-zero $\bar v, span(\bar v)$ is not spanned by any proper subset of $\{ \bar v \}$.
Edit: Definition: Let $S=\{ \bar v_i:i\in I\}$ be a subset of a vector space $V$, then $$span(S):= \{ \sum_{i \in I}c_i\bar v_i: \bar v_i \in S, c_i \in \Bbb F, c_i=0 \mbox { for almost all } i \}.$$ Taking $I=\emptyset$, i.e. $S=\emptyset$, we get $$span(\emptyset )= \{ \sum_{i \in \varnothing }c_i\bar v_i \} = \{ \bar o \},$$ because by definition the empty sum equals $\bar o$ (just like in arithmetic, where the empty sum equals $0$ and the empty product equals $1$, or in set theory, where the empty union equals $\emptyset$).
