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Show that the space \(C\left( \mathbb{R} \right)\) of all continuous functions defined on the real line is infinite-dimensional.

Short Answer

Expert verified

It is proved that the space \(C\left( \mathbb{R} \right)\) of all continuous functions is infinite-dimensional.

Step by step solution

01

Explain finite-dimensional

Suppose \(H\) is a subspace of a finite-dimensional vector space \(V\). Then, according totheorem 11,anylinearly independent setin \(H\) can be expanded, if necessary, to a basis for \(H\). Also, \(H\) is finite-dimensional and \(\dim H \le \dim V\).

02

Show that the space \(C\left( \mathbb{R} \right)\) of all continuous functions is infinite-dimensional

The space \(C\left( \mathbb{R} \right)\) is the subspace of \({\mathop{\rm P}\nolimits} \). According to theorem 11, if the space \(C\left( \mathbb{R} \right)\) is finite-dimensional, then \({\mathop{\rm P}\nolimits} \) must also be finite-dimensional. \(C\left( \mathbb{R} \right)\) must also be infinite-dimensional because \({\mathop{\rm P}\nolimits} \) is infinite-dimensional, according to Exercise 27.

Thus, it is proved that the space \(C\left( \mathbb{R} \right)\) of all continuous functions is infinite-dimensional.

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Most popular questions from this chapter

If A is a \({\bf{4}} \times {\bf{3}}\) matrix, what is the largest possible dimension of the row space of A? If Ais a \({\bf{3}} \times {\bf{4}}\) matrix, what is the largest possible dimension of the row space of A? Explain.

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In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

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