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Chapter 4: Q23Q (page 191)

Let \(B\) be the basis of \({{\mathop{\rm P}\nolimits} _3}\) consisting of the Hermite polynomials in Exercise 21, and let \(p\left( t \right) = 7 - 12t - 8{t^2} + 12{t^3}\). Find the coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\).

Short Answer

Expert verified

The coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\3\\{ - 2}\\{\frac{3}{2}}\end{array}} \right]\).

Step by step solution

01

Definition of the coordinate vector of x

Suppose\(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\)is a basis for\(V\)and x is in\(V\). Thecoordinatesof\({\mathop{\rm x}\nolimits} \)relative to the basis \(B\)(or the\(B\)-coordinates of x) are the weights \({c_1},...,{c_n}\) such that \({\mathop{\rm x}\nolimits} = {c_1}{b_1} + ... + {c_n}{b_n}\).

02

Determine the coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\)

The coordinate vector of\(p\left( t \right) = 7 - 12t - 8{t^2} + 12{t^3}\)with respect to\(B\)is as shown below:

\({c_1}\left( 1 \right) + {c_2}\left( {2t} \right) + {c_3}\left( { - 2 + 4{t^2}} \right) + {c_4}\left( { - 12t + 8{t^3}} \right) = 7 - 12t - 8{t^2} + 12{t^3}\)

Equate the coefficient of\(t\)to produce the system of the equation as shown below:

\(\begin{aligned} {c_1}\,\,\,\,\,\,\,\,\,\, - 2{c_3}\,\,\,\,\,\,\,\,\,\,\,\,\, &= 7\\\,\,\,\,\,\,\,2{c_2}\,\,\,\,\,\,\,\,\,\, - 12{c_4} &= - 12\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{c_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= - 8\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8{c_4} &= 12\end{aligned}\)

By solving the system of the equation, you get\({c_1} = 3,{c_2} = 3,{c_3} = - 2,{c_4} = \frac{3}{2}\). Therefore,\({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\3\\{ - 2}\\{\frac{3}{2}}\end{array}} \right]\).

Thus, the coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\3\\{ - 2}\\{\frac{3}{2}}\end{array}} \right]\).

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Most popular questions from this chapter

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

\({\left( {{\bf{2}} - t} \right)^{\bf{3}}}\), \({\left( {{\bf{3}} - t} \right)^2}\), \({\bf{1}} + {\bf{6}}t - {\bf{5}}{t^{\bf{2}}} + {t^{\bf{3}}}\)

In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right sides of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss.

(M) Let \({{\mathop{\rm a}\nolimits} _1},...,{{\mathop{\rm a}\nolimits} _5}\) denote the columns of the matrix \(A\), where \(A = \left( {\begin{array}{*{20}{c}}5&1&2&2&0\\3&3&2&{ - 1}&{ - 12}\\8&4&4&{ - 5}&{12}\\2&1&1&0&{ - 2}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm a}\nolimits} _1}}&{{{\mathop{\rm a}\nolimits} _2}}&{{{\mathop{\rm a}\nolimits} _4}}\end{array}} \right)\)

  1. Explain why \({{\mathop{\rm a}\nolimits} _3}\) and \({{\mathop{\rm a}\nolimits} _5}\) are in the column space of \(B\).
  2. Find a set of vectors that spans \({\mathop{\rm Nul}\nolimits} A\).
  3. Let \(T:{\mathbb{R}^5} \to {\mathbb{R}^4}\) be defined by \(T\left( x \right) = A{\mathop{\rm x}\nolimits} \). Explain why \(T\) is neither one-to-one nor onto.
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