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Chapter 4: Q30E (page 191)

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

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

Short Answer

Expert verified

The polynomials are linearly dependent.

Step by step solution

01

Write the polynomials in the standard vector form

The vectors of the given polynomials can be written as follows:

\(\begin{array}{c}{\left( {2 - t} \right)^3} = 8 - 12t + 6{t^2} - {t^3}\\ \equiv \left( {\begin{array}{*{20}{c}}8\\{ - 12}\\6\\{ - 1}\end{array}} \right)\end{array}\),

\(\begin{array}{c}{\left( {3 - t} \right)^2} = 9 - 6t + {t^2}\\ \equiv \left( {\begin{array}{*{20}{c}}9\\{ - 6}\\1\\0\end{array}} \right)\end{array}\)

and

\(1 + 6t - 5{t^2} + {t^3} \equiv \left( {\begin{array}{*{20}{c}}1\\6\\{ - 5}\\1\end{array}} \right)\)

02

Form the matrix using the vectors

The matrix formed by using the vectors of the polynomials is:

\(A = \left( {\begin{array}{*{20}{c}}8&9&1\\{ - 12}&{ - 6}&6\\6&1&{ - 5}\\{ - 1}&0&1\end{array}} \right)\)

03

Write the matrix in the echelon form

\(\left( {\begin{array}{*{20}{c}}8&9&1\\{ - 12}&{ - 6}&6\\6&1&{ - 5}\\{ - 1}&0&1\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\0&1&1\\0&0&0\\0&0&0\end{array}} \right)\)

From the echelon form, it can be observed that for three variables, there are two equations. Hence, one free variable is present.

So, the polynomials are linearly dependent.

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Most popular questions from this chapter

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),鈥�.,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

Consider the polynomials , and \({p_{\bf{3}}}\left( t \right) = {\bf{2}}\) \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + t,{p_{\bf{2}}}\left( t \right) = {\bf{1}} - t\)(for all t). By inspection, write a linear dependence relation among \({p_{\bf{1}}},{p_{\bf{2}}},\) and \({p_{\bf{3}}}\). Then find a basis for Span\(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\).

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)

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

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)
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