We shall say, whenever x=\sum_{i} \alpha_{i} x_{i} , that x is a linear combination of \{x_{i}\} ; we shall use without any further explanation all the simple grammatical implications of this terminology. Thus we shall say, in case x is a linear combination of \{x_{i}\} , that x is linearly dependent on \{x_{i}\} ; we shall leave to the reader the proof that if \{x_{i}\} is linearly independent, then a necessary and sufficient condition that x be a linear combination of \{x_{i}\} is that the enlarged set, obtained by adjoining x to \{x_{i}\} , be linearly dependent. Note that, in accordance with the definition of an empty sum, the origin is a linear combination of the empty set of vectors; it is, moreover, the only vector with this property.
The following theorem is the fundamental result concerning linear dependence.
Theorem 1. The set of non-zero vectors x_{1}, \ldots, x_{n} is linearly dependent if and only if some x_{k} , 2 \leq k \leq n , is a linear combination of the preceding ones.
Proof. Let us suppose that the vectors x_{1}, \ldots, x_{n} are linearly dependent, and let k be the first integer between 2 and n for which x_{1}, \ldots, x_{k} are linearly dependent. (If worse comes to worst, our assumption assures us that k=n will do.) Then \alpha_{1} x_{1}+\cdots+\alpha_{k} x_{k}=0 for a suitable set of \alpha ’s (not all zero); moreover, whatever the \alpha ’s, we cannot have \alpha_{k}=0 , for then we should have a linear dependence relation among x_{1}, \ldots, x_{k-1} , contrary to the definition of k . Hence x_{k}=\frac{-\alpha_{1}}{\alpha_{k}} x_{1}+\cdots+\frac{-\alpha_{k-1}}{\alpha_{k}} x_{k-1} as was to be proved. This proves the necessity of our condition; sufficiency is clear since, as we remarked before, every set containing a linearly dependent set is itself such. ◻
