Linear combinations

We shall say, whenever $x = \sum_{i} α_{i} x_{i}$ , 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 $x_{1}, \dots, x_{n}$ 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 $x_{1}, \dots, x_{n}$ are linearly dependent, and let $k$ be the first integer between $2$ and $n$ for which $x_{1}, \dots, x_{k}$ are linearly dependent. (If worse comes to worst, our assumption assures us that $k = n$ will do.) Then $α_{1} x_{1} + \dots + α_{k} x_{k} = 0$ for a suitable set of $α$ ’s (not all zero); moreover, whatever the $α$ ’s, we cannot have $α_{k} = 0$ , for then we should have a linear dependence relation among $x_{1}, \dots, x_{k - 1}$ , contrary to the definition of $k$ . Hence $x_{k} = \frac{- α_{1}}{α_{k}} x_{1} + \dots + \frac{- α_{k - 1}}{α_{k}} x_{k - 1}$ 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. ◻

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