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(M) Show that \(\left\{ {t,sin\,t,cos\,{\bf{2}}t,sin\,t\,cos\,t} \right\}\) is a linearly independent set of functions defined on \(\mathbb{R}\). Start by assuming that

\({c_{\bf{1}}} \cdot t + {c_{\bf{2}}} \cdot sin\,t + {c_{\bf{3}}} \cdot cos\,{\bf{2}}t + {c_{\bf{4}}} \cdot sin\,t\,cos\,t = {\bf{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{5}} \right)\)

Equation (5) must hold for all real t, so choose several specific values of t (say, \(t = {\bf{0}},\,.{\bf{1}},\,.{\bf{2}}\)) until you get a system of enough equations to determine that the \({c_j}\) must be zero.

Short Answer

Expert verified

By the inverse matrix theorem, this system has only a trivial solution. Hence, \(\left\{ {t,\sin t,\cos 2t,\sin t\cos t} \right\}\) is a linearly independent set of functions.

Step by step solution

01

Write the given statement

Assume \({c_1} \cdot t + {c_2} \cdot \sin t + {c_3} \cdot \cos 2t + {c_4} \cdot \sin t\cos t = 0\).

02

Form a system using specific values of t

The above equation gives a system for \(\) as shown below:

\(\left( {\begin{array}{*{20}{c}}0&{\sin 0}&{\cos 0}&{\sin 0\cos 0}\\{.1}&{\sin .1}&{\cos .2}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\cos .4}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\cos .6}&{\sin .3\cos .3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\\{{c_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\)

It means \(Ac = 0\).

Here, \(A = \left( {\begin{array}{*{20}{c}}0&{\sin 0}&{\cos 0}&{\sin 0\cos 0}\\{.1}&{\sin .1}&{\cos .2}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\cos .4}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\cos .6}&{\sin .3\cos .3}\end{array}} \right)\).

03

Find the determinant of A

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}0&0&1&0\\{.1}&{\sin .1}&{\cos .2}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\cos .4}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\cos .6}&{\sin .3\cos .3}\end{array}} \right|\\ = 1\left| {\begin{array}{*{20}{c}}{.1}&{\sin .1}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\sin .3\cos .3}\end{array}} \right|\\\det A \ne 0\end{array}\)

04

Conclusion

By the inverse matrix theorem, the equation \(Ac = 0\) has only a trivial solution.

Hence, \(\left\{ {t,\sin t,\cos 2t,\sin t\cos t} \right\}\) is a linearly independent setof functions.

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Most popular questions from this chapter

Let be a basis of\({\mathbb{R}^n}\). .Produce a description of an \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)matrix A that implements the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\). Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)). (See Exercise 21.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

Question 11: Let\(S\)be a finite minimal spanning set of a vector space\(V\). That is,\(S\)has the property that if a vector is removed from\(S\), then the new set will no longer span\(V\). Prove that\(S\)must be a basis for\(V\).

(calculus required) Define \(T:C\left( {0,1} \right) \to C\left( {0,1} \right)\) as follows: For f in \(C\left( {0,1} \right)\), let \(T\left( t \right)\) be the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) such that \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\). Show that \(T\) is a linear transformation, and describe the kernel of \(T\). (See the notation in Exercise 20 of Section 4.1.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

21. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 2}&{ - 4.2}&{ - 4.8}&{ - 3.6}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

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