A set consisting of a single vector v is linearly dependent if and only if v = 0? Wouldn't any set consisting of any single vector v be linearly dependent because it can be expressed as a linear combination cv where c = 1? 1 times any vector is itself, right? I don't understand why that would only be true if v = 0. 
 A: A set (better, a list) $v_1,v_2,\dots,v_n$ of vectors is linearly dependent if there exist $a_1,a_2,\dots,a_n$ not all zero with $a_1v_1+a_2v_2+\dots+a_nv_n=0$
If $n=1$, you have your statement:


*

*if $v_1=0$, then $a_1=1$ can be chosen so $a_1v_1=0$, making the list consisting of the zero vector linearly dependent;

*if $v_1\ne0$, then $a_1v_1=0$ implies $a_1=0$; thus the list consisting of $v_1\ne0$ is linearly independent.
A: The definition of linear independence of a set of vectors $\{v_1,\dots,v_n\}$ is that if $c_1v_1+\cdots+c_nv_n=0$, then $c_1=\cdots=c_n=0$. When there’s only one vector in the set, this is obviously true: the only way to get the zero vector is to multiply this lone vector by zero.   
This definition is indeed equivalent to saying that the set is linearly dependent if some vector in the set can be written as a linear combination of the other vectors. When there’s only one vector, though, there are no other vectors left over to form this linear combination, so a set consisting of a single vector can’t be linearly dependent.
