Question

Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

Short answer

The set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) is not linearly dependent according to theorem 7; it is linearly independent.

Step-by-step solution

Prove directly that the set is linearly independent

Theorem 7states that an indexed set \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{v_p}} \right\}\) of two or more vectors islinearly dependentif and only if at least one of the vectors in \(S\) is a linear combination of the others. If \(S\) is linearly dependent and \({{\mathop{\rm v}\nolimits} _1} \ne 0\), then some \({{\mathop{\rm v}\nolimits} _j}\) is a linear combination of the preceding vectors \({{\mathop{\rm v}\nolimits} _1},...,{v_{j - 1}}\).

The set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent, and \({{\mathop{\rm v}\nolimits} _1} \ne 0\). Theorem 7 further establishes that \({{\mathop{\rm v}\nolimits} _2}\) cannot be a multiple of \({{\mathop{\rm v}\nolimits} _1}\), and \({{\mathop{\rm v}\nolimits} _3}\) cannot be a linear combination of \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\). According to this hypothesis, \({{\mathop{\rm v}\nolimits} _4}\) is not a linear combination of \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},\) and \({{\mathop{\rm v}\nolimits} _3}\). Thus, \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) is not a linearly dependent set according to theorem 7. So, it must be linearly independent.

Prove that the set is linearly independent by contradiction

Consider \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) as a linearly dependent set. Theorem 7 states that one of the vectors in the set is a linear combination of the preceding vectors. The vector cannot be \({{\mathop{\rm v}\nolimits} _4}\) since \({{\mathop{\rm v}\nolimits} _4}\) is not in the span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\). Furthermore, none of the vectors in \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is a linear combination of the preceding vectors. Therefore, the linear dependence of the set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) is impossible.

Thus, the set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) is linearly independent.