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Question: If \(c \in \mathbb{R}\) and S is a set, define \(cS = \left\{ {c{\bf{x}}:{\bf{x}} \in S} \right\}\). Let S be a convex set and suppose \(c > {\bf{0}}\) and \(d > {\bf{0}}\). Prove that \(cS + dS = \left( {c + d} \right)S\).

Short Answer

Expert verified

It is proved that \(cS + dS = \left( {c + d} \right)S\).

Step by step solution

01

Simplify the expression \(cS + dS\)

As S is a convex set, therefore it can be represented as shown below:

\(\begin{array}{c}cS + dS = \left\{ {cx + dy:\,\,x,y \in S} \right\}\\\left( {c + d} \right)S = \left\{ {\left( {c + d} \right)z:\,z \in S} \right\}\\\left( {c + d} \right)S \subset cS + dS\end{array}\)

If the above equation is true, then \(x = y = z\).

Let, \(z = \frac{c}{{c + d}}x + \frac{d}{{c + d}}y\).

Since c and d both are positive. Therefore z can be represented as aconvex combination of x and y.

02

Prove the given identity

As it was established that S is convex, therefore it can be observed that:

\(\begin{array}{c}\left( {c + d} \right)z \in \left( {c + d} \right)S\\\left( {c + d} \right)z = cz + dz\\\left( {c + d} \right)z \in cS + dS\end{array}\)

Hence, it is proved that \(cS + dS = \left( {c + d} \right)S\).

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Most popular questions from this chapter

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.

Question: In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{array}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{array}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each is given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)

a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{ - {\bf{19}}}\\{\bf{5}}\end{array}} \right)\) b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{{\bf{1}}.{\bf{5}}}\\{ - {\bf{1}}.{\bf{3}}}\\{ - .{\bf{5}}}\end{array}} \right)\) c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{4}}}\\{\bf{0}}\end{array}} \right)\)

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

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\\{ - {\bf{2}}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\\{\bf{6}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{{\bf{17}}}\\{\bf{1}}\\{\bf{5}}\end{aligned}} \right)\)

Question: 13. Suppose \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\). Show that Span \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) is a plane in \({\mathbb{R}^3}\). (Hint: What can you say about \({\rm{u}}\) and \({\rm{v}}\)when Span \(\left\{ {{\rm{u,v}}} \right\}\) is a plane?)

Question 3: Repeat Exercise 1 where \(m\) is the minimum value of f on \(S\) instead of the maximum value.

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