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Use only the definition of affine dependence to show that anindexed set \(\left\{ {{v_1},{v_2}} \right\}\) in \({\mathbb{R}^{\bf{n}}}\) is affinely dependent if and only if \({v_1} = {v_2}\).

Short Answer

Expert verified

An indexed set \(\left\{ {{v_1},{v_2}} \right\}\) in \({\mathbb{R}^n}\) is affinely dependent if and only if \({{\bf{v}}_1} = {{\bf{v}}_2}\).

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

An indexed set \(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\)in\({\mathbb{R}^n}\)is affinely dependent if and only if\({{\bf{v}}_{\bf{1}}} = {{\bf{v}}_{\bf{2}}}\).

The set\(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\)in\({\mathbb{R}^n}\)exists if for non-zero scalars\({c_1}\)and\({c_2}\), the sum is zero i.e.\({c_1} + {c_2} = 0\)and\({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} = 0\).

Here,\({c_1} = - {c_2} \ne 0\). So,\({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} = 0\)can be written as shown below:

\(\begin{aligned}{}\left( { - {c_2}} \right){{\bf{v}}_1} + {c_2}{{\bf{v}}_2} = 0\\ - {c_2}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} = 0\\ - {c_2}\left( {{{\bf{v}}_1} - {{\bf{v}}_2}} \right) = 0\\{{\bf{v}}_1} - {{\bf{v}}_2} = 0\end{aligned}\)

So,\({{\bf{v}}_1} = {{\bf{v}}_2}\).

Thus, an indexed set \(\left\{ {{v_1},{v_2}} \right\}\) in \({\mathbb{R}^n}\) is affinely dependent if and only if \({{\bf{v}}_1} = {{\bf{v}}_2}\).

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Most popular questions from this chapter

Question: In Exercises 5-8, find the minimal representation of the polytope defined by the inequalities \(A{\bf{x}} \le {\bf{b}}\) and \({\bf{x}} \ge {\bf{0}}\).

5. \(A = \left( {\begin{array}{*{20}{c}}1&2\\3&1\end{array}} \right),{\rm{ }}{\bf{b}} = \left( {\begin{array}{*{20}{c}}{{\bf{10}}}\\{{\bf{15}}}\end{array}} \right)\)

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.

13. a. Use equation (12) to show that \({{\mathop{\rm q}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\).

b. Use equation (13) to show that \(8{{\mathop{\rm q}\nolimits} _2} = 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\).

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

Question: 12. In Exercises 11 and 12, mark each statement True or False. Justify each answer.

a. If \(S = \left\{ {\bf{x}} \right\}\), then \({\rm{aff}}\,S\) is the empty set.

b. A set is affine if and only if it contains its affine hull.

c. A flat of dimension 1 is called a line.

d. A flat of dimension 2 is called a hyper plane.

e. A flat through the origin is a subspace.

In Exercises 1-4, write y as an affine combination of the other point listed, if possible.

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{2}}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{7}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{aligned}} \right)\)

Repeat Exercise 25 with\({v_1} = \left[ {\begin{array}{*{20}{c}}1\\{\bf{2}}\\{ - {\bf{4}}}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{8}}\\{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{{\bf{10}}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{8}}\end{array}} \right]\), and \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{.{\bf{9}}}\\{{\bf{2}}.{\bf{0}}}\\{ - {\bf{3}}.{\bf{7}}}\end{array}} \right]\).

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