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Suppose that\(\left\{ {{p_1},{p_2},{p_3}} \right\}\)is an affinely independent set in\({\mathbb{R}^{\bf{n}}}\)and q is an arbitrary point in\({\mathbb{R}^{\bf{n}}}\). Show that the translated set\(\left\{ {{p_1} + q,{p_2} + q,{p_3} + {\bf{q}}} \right\}\)is also affinely independent.

Short Answer

Expert verified

The translated set is also affinely independent.

Step by step solution

01

State the condition for affinely dependence

The set is said to be affinely dependent if the set \(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},...,{{\bf{v}}_p}} \right\}\) in the dimension \({\mathbb{R}^n}\) exists such that for non-zero scalars \({c_1},{c_2},...,{c_p}\), the sum of scalars is zero i.e. \({c_1} + {c_2} + ... + {c_p} = 0\), and \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\).

02

Show affinely dependence 

The set is affinely dependent if the set\(\left\{ {{p_1} + q,{p_2} + q,{p_3} + {\bf{q}}} \right\}\)in\({\mathbb{R}^n}\)exists such that for non-zero scalars\({c_1},{c_2}\)and\({c_3}\)the is zero i.e.\({c_1} + {c_2} + {c_3} = 0\).

So,\({c_1}\left( {{p_1} + q} \right) + {c_2}\left( {{p_2} + q} \right) + {c_3}\left( {{p_3} + q} \right) = 0\).

Simplify\({c_1}\left( {{p_1} + q} \right) + {c_2}\left( {{p_2} + q} \right) + {c_3}\left( {{p_3} + q} \right) = 0\)as shown below:

\(\begin{array}{c}{c_1}{p_1} + {c_1}q + {c_2}{p_2} + {c_2}q + {c_3}{p_3} + {c_3}q = 0\\{c_1}{p_1} + {c_2}{p_2} + {c_3}{p_3} + \left( {{c_1} + {c_2} + {c_3}} \right){\bf{q}} = 0\end{array}\)

As\({c_1} + {c_2} + {c_3} = 0\),\({c_1}{p_1} + {c_2}{p_2} + {c_3}{p_3} = 0\).

Thus,\({c_1}{{\bf{p}}_1} + {c_2}{{\bf{p}}_2} + {c_3}{{\bf{p}}_3} = 0\)and\({c_1} + {c_2} + {c_3} = 0\)show that\(S = \left\{ {{p_1},{p_2},{p_3}} \right\}\) is an affinely dependent set of points in \({\mathbb{R}^n}\).

But\(\left\{ {{p_1},{p_2},{p_3}} \right\}\)is affinely independent. Thus, the translated set is also affinely independent.

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Most popular questions from this chapter

Let \({\bf{x}}\left( t \right)\) be a cubic Bézier curve determined by points \({{\bf{p}}_o}\), \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), and \({{\bf{p}}_3}\).

a. Compute the tangent vector \({\bf{x}}'\left( t \right)\). Determine how \({\bf{x}}'\left( 0 \right)\) and \({\bf{x}}'\left( 1 \right)\) are related to the control points, and give geometric descriptions of the directions of these tangent vectors. Is it possible to have \({\bf{x}}'\left( 1 \right) = 0\)?

b. Compute the second derivative and determine how and are related to the control points. Draw a figure based on Figure 10, and construct a line segment that points in the direction of . [Hint: Use \({{\bf{p}}_1}\) as the origin of the coordinate system.]

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In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

14.a. Justify each equal sign:

\(3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\)

b. Show that \({{\mathop{\rm r}\nolimits} _2}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _2}\) to \({{\mathop{\rm p}\nolimits} _3}\).

c. Justify each equal sign: \(3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right)\).

d. Use part (c) to show that \(8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}\).

e. Use part (d) equation (8), and part (a) to show that \({{\mathop{\rm r}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm r}\nolimits} _2}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\). That is, \({{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).

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