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The conditions for affine dependence are stronger than those for linear dependence, so an affinely dependent set is automatically linearly dependent. Also, a linearly independent set cannot be affinely dependent and therefore must be affinely independent. Construct two linearly dependent indexed sets\({S_{\bf{1}}}\)and\({S_{\bf{2}}}\)in\({\mathbb{R}^2}\)such that\({S_{\bf{1}}}\)is affinely dependent and\({S_{\bf{2}}}\)is affinely independent. In each case, the set should contain either one, two, or three nonzero points.

Short Answer

Expert verified

The indexed set \({S_1}\)is affinelyand linearly dependent.

The indexed set \({S_2}\) is affinelyand linearly 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

Let\({S_1} = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},{{\bf{v}}_3}} \right\}\)be the set of vectors, where\({{\bf{v}}_1} = \left[ {\begin{aligned}{{}{}}3\\{ - 3}\end{aligned}} \right]\),\({{\bf{v}}_2} = \left[ {\begin{aligned}{{}{}}0\\6\end{aligned}} \right]\), and\({{\bf{v}}_3} = \left[ {\begin{aligned}{{}{}}2\\0\end{aligned}} \right]\).

Let the newpoints, \({{\bf{v}}_3} - {{\bf{v}}_1}\)and \({{\bf{v}}_2} - {{\bf{v}}_1}\), be obtained by eliminating the first point.

\(\begin{aligned}{}{{\bf{v}}_3} - {{\bf{v}}_1} &= \left[ {\begin{aligned}{{}{}}2\\0\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}3\\{ - 3}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{ - 1}\\3\end{aligned}} \right]\end{aligned}\)

And

\(\begin{aligned}{}{{\bf{v}}_2} - {{\bf{v}}_1} &= \left[ {\begin{aligned}{{}{}}0\\6\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}3\\{ - 3}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{ - 3}\\9\end{aligned}} \right]\end{aligned}\)

So, \({{\bf{v}}_2} - {{\bf{v}}_1} = 3\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\).

Simplify the equation \({{\bf{v}}_2} - {{\bf{v}}_1} = 3\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\).

\[\begin{aligned}{}{{\bf{v}}_2} - {{\bf{v}}_1} &= 3{{\bf{v}}_3} - 3{{\bf{v}}_1}\\2{{\bf{v}}_1} + {{\bf{v}}_2} - 3{{\bf{v}}_3} &= 0\end{aligned}\]

If the vectors are affinely dependent, the sum of weightsis 0. In the equation \(3{{\bf{v}}_1} + {{\bf{v}}_2} - 3{{\bf{v}}_3} = 0\), the weights are 2, 1, and \( - 3\). So,

\(2 + 1 - 3 = 0\).

Thus, the indexed set \({S_1}\) is affinelyand linearly dependent.

03

 Step 3: Show affinely independence

Let\({S_2} = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\)be the set of vectors, where\({{\bf{v}}_1} = \left[ {\begin{aligned}{{}{}}1\\2\end{aligned}} \right]\)and\({{\bf{v}}_2} = \left[ {\begin{aligned}{{}{}}2\\4\end{aligned}} \right]\).

So, \({{\bf{v}}_2} = 2{{\bf{v}}_1}\).

Simplify the equation \({{\bf{v}}_2} = 2{{\bf{v}}_1}\).

\({{\bf{v}}_2} - 2{{\bf{v}}_1} = 0\)

For thevectors to be affinely dependent, the sum of weights must be 0. In the equation \({{\bf{v}}_2} - 2{{\bf{v}}_1} = 0\), the weights are1 and \( - 2\). So,

\(\begin{aligned}{}1 - 2 &= - 1\\ &\ne 0\end{aligned}\).

Thus, the indexed set \({S_2}\) is affinelyand linearly independent.

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Most popular questions from this chapter

Question: In Exercises 15-20, write a formula for a linear functional f and specify a number d, so that \(\left) {f:d} \right)\) the hyperplane H described in the exercise.

Let H be the plane in \({\mathbb{R}^{\bf{3}}}\) spanned by the rows of \(B = \left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{4}}&{ - {\bf{5}}}\\{\bf{0}}&{ - {\bf{2}}}&{\bf{8}}\end{array}} \right)\). That is, \(H = {\bf{Row}}\,B\).

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

11.a. The cubic Bezier curve is based on four control points.

b. Given a quadratic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\), the directed line segment \({{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}\) (from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\)) is the tangent vector to the curve at \({{\mathop{\rm p}\nolimits} _0}\).

c. When two quadratic Bezier curves with control points \(\left\{ {{{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2}} \right\}\) and \(\left\{ {{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3},{{\mathop{\rm p}\nolimits} _4}} \right\}\) are joined at \({{\mathop{\rm p}\nolimits} _2}\), the combined Bezier curve will have \({C^1}\) continuity at \({{\mathop{\rm p}\nolimits} _2}\)if\({{\mathop{\rm p}\nolimits} _2}\) is the midpoint of the line segment between \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _3}\).

In Exercises 7 and 8, find the barycentric coordinates of p with respect to the affinely independent set of points that precedes it.

8. \(\left( {\begin{array}{{}}0\\1\\{ - 2}\\1\end{array}} \right),\left( {\begin{array}{{}}1\\1\\0\\2\end{array}} \right),\left( {\begin{array}{{}}1\\4\\{ - 6}\\5\end{array}} \right)\), \({\mathop{\rm p}\nolimits} = \left( {\begin{array}{{}}{ - 1}\\1\\{ - 4}\\0\end{array}} \right)\)

In Exercises 21–24, a, b, and c are noncollinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \)and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

23. Let p be any point in the interior of\(\Delta {\bf{abc}}\), with barycentric coordinates\(\left( {r,s,t} \right)\), so that

\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}r\\s\\t\end{array}} \right] = \widetilde {\bf{p}}\)

Use Exercise 21 and a fact about determinants (Chapter 3) to show that

\(r = \left( {area of \Delta pbc} \right)/\left( {area of \Delta abc} \right)\)

\(s = \left( {area of \Delta apc} \right)/\left( {area of \Delta abc} \right)\)

\(t = \left( {area of \Delta abp} \right)/\left( {area of \Delta abc} \right)\)

Let \(W = \left\{ {{{\bf{v}}_1},......,{{\bf{v}}_p}} \right\}\). Show that if \({\bf{x}}\) is orthogonal to each \({{\bf{v}}_j}\), for \(1 \le j \le p\), then \({\bf{x}}\) is orthogonal to every vector in \(W\).

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