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=== The zero vector === If one or more vectors from a given sequence of vectors <math>\mathbf{v}_1, \dots, \mathbf{v}_k</math> is the zero vector <math>\mathbf{0}</math> then the vectors <math>\mathbf{v}_1, \dots, \mathbf{v}_k</math> are necessarily linearly dependent (and consequently, they are not linearly independent). To see why, suppose that <math>i</math> is an index (i.e. an element of <math>\{ 1, \ldots, k \}</math>) such that <math>\mathbf{v}_i = \mathbf{0}.</math> Then let <math>a_{i} := 1</math> (alternatively, letting <math>a_{i}</math> be equal to any other non-zero scalar will also work) and then let all other scalars be <math>0</math> (explicitly, this means that for any index <math>j</math> other than <math>i</math> (i.e. for <math>j \neq i</math>), let <math>a_{j} := 0</math> so that consequently <math>a_{j} \mathbf{v}_j = 0 \mathbf{v}_j = \mathbf{0}</math>). Simplifying <math>a_1 \mathbf{v}_1 + \cdots + a_k\mathbf{v}_k</math> gives: :<math>a_1 \mathbf{v}_1 + \cdots + a_k\mathbf{v}_k = \mathbf{0} + \cdots + \mathbf{0} + a_i \mathbf{v}_i + \mathbf{0} + \cdots + \mathbf{0} = a_i \mathbf{v}_i = a_i \mathbf{0} = \mathbf{0}.</math> Because not all scalars are zero (in particular, <math>a_{i} \neq 0</math>), this proves that the vectors <math>\mathbf{v}_1, \dots, \mathbf{v}_k</math> are linearly dependent. As a consequence, the zero vector can not possibly belong to any collection of vectors that is linearly ''in''dependent. Now consider the special case where the sequence of <math>\mathbf{v}_1, \dots, \mathbf{v}_k</math> has length <math>1</math> (i.e. the case where <math>k = 1</math>). A collection of vectors that consists of exactly one vector is linearly dependent if and only if that vector is zero. Explicitly, if <math>\mathbf{v}_1</math> is any vector then the sequence <math>\mathbf{v}_1</math> (which is a sequence of length <math>1</math>) is linearly dependent if and only if {{nowrap|<math>\mathbf{v}_1 = \mathbf{0}</math>;}} alternatively, the collection <math>\mathbf{v}_1</math> is linearly independent if and only if <math>\mathbf{v}_1 \neq \mathbf{0}.</math>
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