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Chapter 1: Q18Q (page 1)

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.

Short Answer

Expert verified

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

Step by step solution

01

Consider the vector equation

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

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

02

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

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.

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If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

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