A question on conditions of linear independence Alright, so I've learned the three basic criteria that make a set of vectors linearly dependent. I'll go ahead and list them.

*

*More vectors than entries per vector, e.g.:
$$\pmatrix{1\\2},\pmatrix{3\\4},\pmatrix{5\\6}$$


*All vectors are scalar multiples of each other, e.g.:
$$\pmatrix{1\\-2\\-5},\pmatrix{-2\\4\\10},\pmatrix{3\\-6\\-15}$$


*Presence of a zero vector $\bar{0}$, e.g.:
$$\pmatrix{1\\2},\pmatrix{0\\0}$$
Clearly determining linear dependence is not too tall a task considering we only have three tests to apply to our set.
However, rather than perform row operations and convert a corresponding matrix to reduced echelon form to determine linear independence, could we say that a set of vectors that fails to pass these three tests is necessarily linearly independent? If not, could someone provide a counterexample set that is dependent but meets none of the above criteria?
 A: The criteria you mention are only special cases.  The set $\left\{\pmatrix{1 \\ 0 \\ 0}, \pmatrix{0 \\ 1 \\ 1}, \pmatrix{1 \\ 1 \\ 1}\right\}$ is linearly dependent but doesn't satisfy any of your criteria.
The actual definition of linear independence is the following:

A set of vectors $\{\mathbf v_1, \dots, \mathbf v_n\}$ is linearly independent if $$a_1\mathbf v_1 + a_2\mathbf v_2 + \cdots + a_n\mathbf v_n = \mathbf 0 \implies a_1 = a_2 = \cdots = a_n = 0$$ where $a_1, \dots, a_n$ are scalars (regular numbers). A set of vectors is linearly dependent if it is not linearly independent.

A necessary and sufficient criterion for linear independence is that none of the vectors are a linear combination of any (finite) subset of the others.  The set I give above fails this test because the last vector is a linear combination of the first two.
A: A set of vectors vectors $\{a_1, \ldots, a_k\}$ is linearly independent if it is not linearly dependent, i.e.,
$$
\beta_1 a_1 + \ldots + \beta_k a_k = 0
$$
holds only when $\beta_1 = \cdots = \beta_k = 0$. 
The criteria you specified are only sufficient. 
