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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

Let S be a maximal linearly independent subset of a vector space V. In other words, S has the property that if a vector not in S is adjoined to S, the new set will no longer be linearly independent. Prove that S must be a basis of V. [Hint: What if S were linearly independent but not a basis of V?]

In Exercise 2, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

2. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{5}}}\end{array}} \right)\)

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

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

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.

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)

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