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Chapter 4: Q14Q (page 191)

The set \(B = \left\{ {1 - {t^2},t - {t^2},2 - 2t + {t^2}} \right\}\) is a basis for \({{\mathop{\rm P}\nolimits} _2}\). Find the coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) relative to \(B\).

Answer:

The coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}7\\{ - 3}\\{ - 2}\end{array}} \right]\).

Short Answer

Expert verified

Step 1: Equate the coefficients of the two polynomials

Step by step solution

01

Equate the coefficients of the two polynomials

Determine \({c_1},{c_2}\), and \({c_3}\) as shown below:

\[{c_1}\left( {1 - {t^2}} \right) + {c_2}\left( {t - {t^2}} \right) + {c_3}\left( {2 - 2t + {t^2}} \right) = {\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\]

Equate the coefficients of the two polynomials to obtain the system of equations as shown below:

\[\begin{array}{c}{c_1} + \,\,\,\,\,\,\,2{c_3} = 3\\\,\,\,\,\,\,\,\,{c_2} - 2{c_3} = 1\\ - {c_1} - {c_2} + {c_3} = - 6\end{array}\]

02

Convert the system of equations into an augmented matrix

Convert the system of equations into an augmented matrix, as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&0&2&3\\0&1&{ - 2}&1\\{ - 1}&{ - 1}&1&{ - 6}\end{array}} \right]\)

03

Apply the row operation

At row 3, add rows 1 and 3.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&2&3\\0&1&{ - 2}&1\\0&{ - 1}&3&{ - 3}\end{array}} \right]\)

At row 3, add rows 2 and 3.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&2&3\\0&1&{ - 2}&1\\0&0&1&{ - 2}\end{array}} \right]\)

At row 1, multiply row 3 by \(2\) and subtract it from row 1.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&7\\0&1&{ - 2}&1\\0&0&1&{ - 2}\end{array}} \right]\)

At row 2, multiply row 3 by 2 and add it to row 2.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&7\\0&1&0&{ - 3}\\0&0&1&{ - 2}\end{array}} \right]\)

04

Determine the coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\)

The coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}7\\{ - 3}\\{ - 2}\end{array}} \right]\).

It is also possible to solve this problem using the coordinate vectors of the given polynomials relative to the standard basis \(\left\{ {1,t,{t^2}} \right\}\). The same system of equations will be obtained in the result.

Thus, the coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}7\\{ - 3}\\{ - 2}\end{array}} \right]\).

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Most popular questions from this chapter

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

(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.)

Let \(V\) and \(W\) be vector spaces, and let \(T:V \to W\) be a linear transformation. Given a subspace \(U\) of \(V\), let \(T\left( U \right)\) denote the set of all images of the form \(T\left( {\mathop{\rm x}\nolimits} \right)\), where x is in \(U\). Show that \(T\left( U \right)\) is a subspace of \(W\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

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

17. A submatrix of a matrix A is any matrix that results from deleting some (or no) rows and/or columns of A. It can be shown that A has rank \(r\) if and only if A contains an invertible \(r \times r\) submatrix and no longer square submatrix is invertible. Demonstrate part of this statement by explaining (a) why an \(m \times n\) matrix A of rank \(r\) has an \(m \times r\) submatrix \({A_1}\) of rank \(r\), and (b) why \({A_1}\) has an invertible \(r \times r\) submatrix \({A_2}\).

The concept of rank plays an important role in the design of engineering control systems, such as the space shuttle system mentioned in this chapter’s introductory example. A state-space model of a control system includes a difference equation of the form

\({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\)for \(k = 0,1,....\) (1)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) is a sequence of “state vectors” in \({\mathbb{R}^n}\) that describe the state of the system at discrete times, and \(\left\{ {{{\mathop{\rm u}\nolimits} _k}} \right\}\) is a control, or input, sequence. The pair \(\left( {A,B} \right)\) is said to be controllable if

\({\mathop{\rm rank}\nolimits} \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\) (2)

The matrix that appears in (2) is called the controllability matrix for the system. If \(\left( {A,B} \right)\) is controllable, then the system can be controlled, or driven from the state 0 to any specified state \({\mathop{\rm v}\nolimits} \) (in \({\mathbb{R}^n}\)) in at most \(n\) steps, simply by choosing an appropriate control sequence in \({\mathbb{R}^m}\). This fact is illustrated in Exercise 18 for \(n = 4\) and \(m = 2\). For a further discussion of controllability, see this text’s website (Case study for Chapter 4).
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