Questions: Let \({F_{\bf{1}}}\) and \({F_{\bf{2}}}\) be 4-dimensional flats in \({\mathbb{R}^{\bf{6}}}\), and suppose that \({F_{\bf{1}}} \cap {F_{\bf{2}}} \ne \phi \). What are the possible dimension of \({F_{\bf{1}}} \cap {F_{\bf{2}}}\)?

Short Answer

Expert verified

The dimension D of the solution set is, \(2 \le D \le 4\).

Step by step solution

01

Find the dimension of \({F_{\bf{1}}} \cap {F_{\bf{2}}}\) for complete overlap

\({F_1}\)and \({F_2}\) have complete overlap, then \({F_1}\), \({F_2}\) have four dimensions.

02

Find the range of dimension

As there is six-dimensional space, at the minimum, they have \(\left( {6 - 4 = 2} \right)\) dimensions. So, the range of dimensions is shown below:

\(2 \le D \le 4\)

Thus, the dimension of the solution set is from 2 to 4.

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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 \(A{\mathop{\rm x}\nolimits} \le {\mathop{\rm b}\nolimits} \) and \({\mathop{\rm x}\nolimits} \ge 0\).

6. \(A = \left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{array}{*{20}{c}}{18}\\{16}\end{array}} \right)\)

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 \({\bf{x}}\left( t \right)\) 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.]

A quartic Bézier curve is determined by five control points,

\({{\bf{p}}_{\bf{o}}}{\bf{,}}\,{\rm{ }}{{\bf{p}}_{\bf{1}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{2}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{3}}}\)and \({{\bf{p}}_4}\):

\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^4}{{\bf{p}}_0} + 4t{\left( {1 - t} \right)^3}{{\bf{p}}_1} + 6{t^2}{\left( {1 - t} \right)^2}{{\bf{p}}_2} + 4{t^3}\left( {1 - t} \right){{\bf{p}}_3} + {t^4}{{\bf{p}}_4}\)for \(0 \le t \le 1\)

Construct the quartic basis matrix \({M_B}\) for \({\bf{x}}\left( t \right)\).

Question: Repeat Exercise 7 when

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{3}}\\{ - {\bf{2}}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{6}}\\{ - {\bf{5}}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{0}}\\{{\bf{12}}}\\{ - {\bf{6}}}\end{array}} \right)\)

\({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{1}}}\\{{\bf{15}}}\\{ - {\bf{7}}}\end{array}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{5}}}\\{\bf{3}}\\{ - {\bf{8}}}\\{\bf{6}}\end{array}} \right)\), and \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{6}}\\{ - {\bf{6}}}\\{ - {\bf{8}}}\end{array}} \right)\)

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