Question

Each statement in Exercises 33–38 is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexampleto the statement. If a statement is true, give a justification. (One specific example cannot explain why a statement is always true. You will have to do more work here than in Exercises 21 and 22.)

38. If \({{\bf{v}}_1},...,{{\bf{v}}_4}\) are linearly independent vectors in \({\mathbb{R}^4}\), then \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is also linearly independent. [Hint:Think about \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + 0 \cdot {{\bf{v}}_4} = 0\).]

Short answer

The statement is true.

Step-by-step solution

Write the condition for the linear independence of the vectors

The vectors are said to be linearly independent if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + ... + {x_p}{{\bf{v}}_p} = 0\) has a trivial solution, where \({{\bf{v}}_1},{{\bf{v}}_2},...,{{\bf{v}}_p}\) are the vectors.

Check whether the statement is true or false

It is given that \({{\bf{v}}_1},...,{{\bf{v}}_4}\)arelinearly independent vectors in \({\mathbb{R}^4}\).

Consider the equation for linear dependence \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + {x_4}{{\bf{v}}_4} = 0\).

Substitute 0 for \({x_4}\) in the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + {x_4}{{\bf{v}}_4} = 0\).

\(\begin{aligned}{c}{x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + 0 \cdot {{\bf{v}}_4} &= 0\\{x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} &= 0\end{aligned}\)

So, it shows that vectors \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) are also linearly independent.

Thus, the statement is true.