Suppose \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) is a linearly independent set in \({\mathbb{R}^n}\). Show that \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _1} + {{\mathop{\rm v}\nolimits} _2}} \right\}\) is also linearly independent.

Let \({c_1}\) and \({c_2}\) be constants such that

\(\begin{aligned}{l}{c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}\left( {{{\mathop{\rm v}\nolimits} _1} + {{\mathop{\rm v}\nolimits} _2}} \right) = 0....\left( * \right)\\\left( {{c_1} + {c_2}} \right){{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = 0.\end{aligned}\)

Vectors \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) are linearly independent because \({c_1} + {c_2} = 0\) and \({c_2} = 0\). Both \({c_1}\) and \({c_2}\) in equation (*) must be zero, which means that \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _1} + {{\mathop{\rm v}\nolimits} _2}} \right\}\) is linearly independent.

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