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Chapter 4: Q37E (page 191)

(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

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

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

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

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

\({\bf{1}} - {\bf{2}}{t^{\bf{2}}} - {t^{\bf{3}}}\), \(t + {\bf{2}}{t^{\bf{3}}}\), \({\bf{1}} + t - {\bf{2}}{t^{\bf{2}}}\)

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: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)
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