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(M) Show thatis a linearly independent set of functions defined on. Use the method of Exercise 37. (This result will be needed in Exercise 34 in Section 4.5.)

Short Answer

Expert verified

By the invertible matrix theorem, is a linearly independent set of functions defined on .

Step by step solution

01

Use the method used in Exercise 37

Assume \({c_1} \cdot 1 + {c_2} \cdot \cos t + {c_3} \cdot {\cos ^2}t + ... + {c_7} \cdot {\cos ^6}t = 0\).

For \(t = 0,.1,.2,...,.6\), the above equation gives the system as shown below:

\(\left( {\begin{array}{*{20}{c}}1&{\cos 0}&{{{\cos }^2}0}&{{{\cos }^3}0}&{{{\cos }^4}0}&{{{\cos }^5}0}&{{{\cos }^6}0}\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\\{{c_4}}\\{{c_5}}\\{{c_6}}\\{{c_7}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\\0\\0\\0\end{array}} \right)\)

That is, \(Ac = 0\).

02

Find the determinant of A

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&{\cos 0}&{{{\cos }^2}0}&{{{\cos }^3}0}&{{{\cos }^4}0}&{{{\cos }^5}0}&{{{\cos }^6}0}\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right|\\\det A \ne 0\end{array}\)

03

Draw a conclusion

By the invertible matrix theorem, has only a trivial solution. Thus, is a linearly independent set of functions defined on .

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Most popular questions from this chapter

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.)

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 \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

The null space of a \({\bf{5}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A.

(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.)

Define by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right)\). For instance, if \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + 5t + 7{t^2}\), then \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}3\\{15}\end{array}} \right)\).

  1. Show that \(T\) is a linear transformation. (Hint: For arbitrary polynomials p, q in \({{\mathop{\rm P}\nolimits} _2}\), compute \(T\left( {{\mathop{\rm p}\nolimits} + {\mathop{\rm q}\nolimits} } \right)\) and \(T\left( {c{\mathop{\rm p}\nolimits} } \right)\).)
  2. Find a polynomial p in \({{\mathop{\rm P}\nolimits} _2}\) that spans the kernel of \(T\), and describe the range of \(T\).
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